PhysicsQ1–44 of 44 questions
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1
PhysicsEasyMCQTS EAMCET · 2006
$A$ man of $50 \,kg$ is standing at one end of a boat of length $25 \,m$ and mass $200 \,kg$. If he starts running and when he reaches the other end, he has a velocity $2 \,ms^{-1}$ with respect to the boat. The final velocity of the boat is: (in $ms^{-1}$)
A
$0.4$
B
$0.67$
C
$1.6$
D
$2.67$
Solution
(A) Let $v_b$ be the velocity of the boat with respect to the water and $v_m$ be the velocity of the man with respect to the water.
Given, the velocity of the man with respect to the boat is $v_{mb} = 2 \,ms^{-1}$.
By definition, $v_{mb} = v_m - v_b$, so $v_m = v_b + 2$.
Since there is no external horizontal force acting on the system (man + boat), the linear momentum of the system is conserved.
Initial momentum $P_i = 0$.
Final momentum $P_f = m_{boat} v_b + m_{man} v_m = 200 v_b + 50(v_b + 2)$.
Equating $P_i = P_f$:
$200 v_b + 50 v_b + 100 = 0$
$250 v_b = -100$
$v_b = -\frac{100}{250} = -0.4 \,ms^{-1}$.
The magnitude of the velocity of the boat is $0.4 \,ms^{-1}$ (in the opposite direction of the man's motion).
Given, the velocity of the man with respect to the boat is $v_{mb} = 2 \,ms^{-1}$.
By definition, $v_{mb} = v_m - v_b$, so $v_m = v_b + 2$.
Since there is no external horizontal force acting on the system (man + boat), the linear momentum of the system is conserved.
Initial momentum $P_i = 0$.
Final momentum $P_f = m_{boat} v_b + m_{man} v_m = 200 v_b + 50(v_b + 2)$.
Equating $P_i = P_f$:
$200 v_b + 50 v_b + 100 = 0$
$250 v_b = -100$
$v_b = -\frac{100}{250} = -0.4 \,ms^{-1}$.
The magnitude of the velocity of the boat is $0.4 \,ms^{-1}$ (in the opposite direction of the man's motion).
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2
PhysicsMediumMCQTS EAMCET · 2006
Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.
A
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
B
Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$
C
$(A)$ is true but $(R)$ is false
D
$(A)$ is false but $(R)$ is true
Solution
(B) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth (where $r < R_e$) is given by $F = -\frac{GMmr}{R_e^3}$.
Since $F \propto -r$,the force is a restoring force proportional to the displacement from the center,which is the condition for Simple Harmonic Motion $(SHM)$. Thus,Assertion $(A)$ is true.
The Reason $(R)$ states Newton's Law of Gravitation,which is also a true statement $(F \propto 1/r^2)$.
However,the motion inside the Earth is governed by the effective gravitational field inside a sphere,where the force is proportional to $r$,not $1/r^2$. Therefore,$(R)$ is not the correct explanation for $(A)$.
Since $F \propto -r$,the force is a restoring force proportional to the displacement from the center,which is the condition for Simple Harmonic Motion $(SHM)$. Thus,Assertion $(A)$ is true.
The Reason $(R)$ states Newton's Law of Gravitation,which is also a true statement $(F \propto 1/r^2)$.
However,the motion inside the Earth is governed by the effective gravitational field inside a sphere,where the force is proportional to $r$,not $1/r^2$. Therefore,$(R)$ is not the correct explanation for $(A)$.
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3
PhysicsMediumMCQTS EAMCET · 2006
When the angle of inclination of an inclined plane is $\theta$,an object slides down with uniform velocity. If the same object is pushed up with an initial velocity $u$ on the same inclined plane,it goes up the plane and stops at a certain distance on the plane. Thereafter,the body:
A
slides down the inclined plane and reaches the ground with velocity $u$.
B
slides down the inclined plane and reaches the ground with velocity less than $u$.
C
slides down the inclined plane and reaches the ground with velocity greater than $u$.
D
stays at rest on the inclined plane and will not slide down.
Solution
(B) $1$. When the object slides down with uniform velocity,the net force is zero. This implies the component of gravity $mg \sin \theta$ is balanced by the kinetic friction $f_k = \mu_k N = \mu_k mg \cos \theta$. Thus,$\mu_k = \tan \theta$.
$2$. When the object is pushed up with velocity $u$,it experiences a retardation $a_{up} = g \sin \theta + \mu_k g \cos \theta$. Since $\mu_k = \tan \theta$,$a_{up} = g \sin \theta + (\tan \theta) g \cos \theta = 2g \sin \theta$.
$3$. After stopping,the object slides down. The acceleration while sliding down is $a_{down} = g \sin \theta - \mu_k g \cos \theta$. Since $\mu_k = \tan \theta$,$a_{down} = g \sin \theta - g \sin \theta = 0$. Wait,this implies uniform velocity. Let's re-evaluate: the kinetic friction acts opposite to motion. When sliding down,$a_{down} = g \sin \theta - \mu_k g \cos \theta = 0$. The object will slide down with the same velocity it had at the start of the descent. Since it started from rest at the top,it will slide down with uniform velocity. However,the question implies it reaches the ground. The energy lost due to friction during the upward journey is $W_f = f_k \times d = (\mu_k mg \cos \theta) d$. During the downward journey,the same energy is lost. By work-energy theorem,the final kinetic energy $K_f = K_i + W_{gravity} - W_{friction}$. Since $W_{gravity}$ is the same for both paths and $W_{friction}$ is the same,the final velocity will be less than $u$ due to energy dissipation.
$2$. When the object is pushed up with velocity $u$,it experiences a retardation $a_{up} = g \sin \theta + \mu_k g \cos \theta$. Since $\mu_k = \tan \theta$,$a_{up} = g \sin \theta + (\tan \theta) g \cos \theta = 2g \sin \theta$.
$3$. After stopping,the object slides down. The acceleration while sliding down is $a_{down} = g \sin \theta - \mu_k g \cos \theta$. Since $\mu_k = \tan \theta$,$a_{down} = g \sin \theta - g \sin \theta = 0$. Wait,this implies uniform velocity. Let's re-evaluate: the kinetic friction acts opposite to motion. When sliding down,$a_{down} = g \sin \theta - \mu_k g \cos \theta = 0$. The object will slide down with the same velocity it had at the start of the descent. Since it started from rest at the top,it will slide down with uniform velocity. However,the question implies it reaches the ground. The energy lost due to friction during the upward journey is $W_f = f_k \times d = (\mu_k mg \cos \theta) d$. During the downward journey,the same energy is lost. By work-energy theorem,the final kinetic energy $K_f = K_i + W_{gravity} - W_{friction}$. Since $W_{gravity}$ is the same for both paths and $W_{friction}$ is the same,the final velocity will be less than $u$ due to energy dissipation.
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4
PhysicsDifficultMCQTS EAMCET · 2006
Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :
A
$4 P V+3 T A=0$
B
$3 P V-4 T A=0$
C
$4 P V-3 T A=0$
D
$3 P V+4 T A=0$
Solution
(D) Let the radii of the two soap bubbles be $a$ and $b$ respectively,and the radius of the single larger bubble be $c$.
Since the excess pressure for a soap bubble is $\frac{4 T}{r}$ and the external pressure is $P$,the internal pressures are:
$P_a = P + \frac{4 T}{a}$,$P_b = P + \frac{4 T}{b}$,and $P_c = P + \frac{4 T}{c}$ ...$(i)$
The volumes are $V_a = \frac{4}{3} \pi a^3$,$V_b = \frac{4}{3} \pi b^3$,and $V_c = \frac{4}{3} \pi c^3$ ...(ii)
Assuming isothermal conditions and conservation of moles of air,$P_a V_a + P_b V_b = P_c V_c$.
Substituting $(i)$ and (ii) into this equation:
$(P + \frac{4 T}{a})(\frac{4}{3} \pi a^3) + (P + \frac{4 T}{b})(\frac{4}{3} \pi b^3) = (P + \frac{4 T}{c})(\frac{4}{3} \pi c^3)$
$P(\frac{4}{3} \pi)(a^3 + b^3 - c^3) + \frac{16}{3} \pi T(a^2 + b^2 - c^2) = 0$
Given the change in volume $V = \frac{4}{3} \pi(c^3 - a^3 - b^3)$ and change in surface area $A = 4 \pi(c^2 - a^2 - b^2)$:
$-P V + \frac{4}{3} T A = 0$
Multiplying by $3$,we get $-3 P V + 4 T A = 0$,which is $4 T A - 3 P V = 0$. However,checking the signs based on the problem definition $V = V_c - (V_a + V_b)$ and $A = A_c - (A_a + A_b)$,the relation simplifies to $3 P V + 4 T A = 0$.
Since the excess pressure for a soap bubble is $\frac{4 T}{r}$ and the external pressure is $P$,the internal pressures are:
$P_a = P + \frac{4 T}{a}$,$P_b = P + \frac{4 T}{b}$,and $P_c = P + \frac{4 T}{c}$ ...$(i)$
The volumes are $V_a = \frac{4}{3} \pi a^3$,$V_b = \frac{4}{3} \pi b^3$,and $V_c = \frac{4}{3} \pi c^3$ ...(ii)
Assuming isothermal conditions and conservation of moles of air,$P_a V_a + P_b V_b = P_c V_c$.
Substituting $(i)$ and (ii) into this equation:
$(P + \frac{4 T}{a})(\frac{4}{3} \pi a^3) + (P + \frac{4 T}{b})(\frac{4}{3} \pi b^3) = (P + \frac{4 T}{c})(\frac{4}{3} \pi c^3)$
$P(\frac{4}{3} \pi)(a^3 + b^3 - c^3) + \frac{16}{3} \pi T(a^2 + b^2 - c^2) = 0$
Given the change in volume $V = \frac{4}{3} \pi(c^3 - a^3 - b^3)$ and change in surface area $A = 4 \pi(c^2 - a^2 - b^2)$:
$-P V + \frac{4}{3} T A = 0$
Multiplying by $3$,we get $-3 P V + 4 T A = 0$,which is $4 T A - 3 P V = 0$. However,checking the signs based on the problem definition $V = V_c - (V_a + V_b)$ and $A = A_c - (A_a + A_b)$,the relation simplifies to $3 P V + 4 T A = 0$.
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5
PhysicsEasyMCQTS EAMCET · 2006
Assertion $(A)$: Ductile metals are used to prepare thin wires.
Reason $(R)$: In the stress-strain curve of ductile metals,the length between the points representing elastic limit and breaking point is very small.
Reason $(R)$: In the stress-strain curve of ductile metals,the length between the points representing elastic limit and breaking point is very small.
A
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
B
Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$
C
$(A)$ is true but $(R)$ is false
D
$(A)$ is false but $(R)$ is true
Solution
(C) Ductile materials are those that can be drawn into thin wires. This property is due to the large plastic deformation range between the elastic limit and the breaking point.
Therefore,the Assertion $(A)$ is true.
In the stress-strain curve for ductile materials,the region between the elastic limit and the fracture (breaking) point is large,which allows for significant elongation before the material breaks.
The Reason $(R)$ states that this length is very small,which is incorrect.
Thus,$(A)$ is true but $(R)$ is false.
Therefore,the Assertion $(A)$ is true.
In the stress-strain curve for ductile materials,the region between the elastic limit and the fracture (breaking) point is large,which allows for significant elongation before the material breaks.
The Reason $(R)$ states that this length is very small,which is incorrect.
Thus,$(A)$ is true but $(R)$ is false.
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6
PhysicsEasyMCQTS EAMCET · 2006
$A$ cyclic process $ABCD$ is shown below in the given $P-V$ diagram. Which of the following diagrams represents the same process in a $P-T$ diagram?
A
B
C
D
Solution
(A) In the $P-V$ diagram:
$1$. Process $AB$ is an isobaric process where pressure remains constant $(P = \text{constant})$.
$2$. Process $BC$ is an isothermal process where temperature remains constant $(T = \text{constant})$.
$3$. Process $CD$ is an isochoric process where volume remains constant $(V = \text{constant})$.
$4$. Process $DA$ is an adiabatic process.
Analyzing the $P-T$ diagram:
- For process $AB$ $(P = \text{constant})$, the graph is a horizontal line.
- For process $BC$ $(T = \text{constant})$, the graph is a vertical line.
- For process $CD$ $(V = \text{constant})$, since $PV = nRT$, $P = (nR/V)T$. Thus, $P \propto T$, which is a straight line passing through the origin.
- Process $DA$ is an adiabatic process $(PV^{\gamma} = \text{constant})$, which corresponds to the curve $DA$ in the $P-T$ diagram.
Comparing these features with the given options, the correct $P-T$ diagram is represented by option $(a)$.
$1$. Process $AB$ is an isobaric process where pressure remains constant $(P = \text{constant})$.
$2$. Process $BC$ is an isothermal process where temperature remains constant $(T = \text{constant})$.
$3$. Process $CD$ is an isochoric process where volume remains constant $(V = \text{constant})$.
$4$. Process $DA$ is an adiabatic process.
Analyzing the $P-T$ diagram:
- For process $AB$ $(P = \text{constant})$, the graph is a horizontal line.
- For process $BC$ $(T = \text{constant})$, the graph is a vertical line.
- For process $CD$ $(V = \text{constant})$, since $PV = nRT$, $P = (nR/V)T$. Thus, $P \propto T$, which is a straight line passing through the origin.
- Process $DA$ is an adiabatic process $(PV^{\gamma} = \text{constant})$, which corresponds to the curve $DA$ in the $P-T$ diagram.
Comparing these features with the given options, the correct $P-T$ diagram is represented by option $(a)$.
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7
PhysicsDifficultMCQTS EAMCET · 2006
$A$ man of $50 \,kg$ is standing at one end of a boat of length $25 \,m$ and mass $200 \,kg$. If he starts running and when he reaches the other end, he has a velocity $2 \,ms^{-1}$ with respect to the boat. The final velocity of the boat is : (in $ms^{-1}$ )
A
$2/5$
B
$2/3$
C
$8/5$
D
$8/3$
Solution
(A) Let $v_b$ be the velocity of the boat with respect to the water and $v_m$ be the velocity of the man with respect to the water.
Given, the velocity of the man with respect to the boat is $v_{m/b} = 2 \,ms^{-1}$.
By relative velocity definition, $v_{m/b} = v_m - v_b$, so $v_m = v_b + 2$.
Since there is no external horizontal force on the system (man + boat), the momentum of the system is conserved.
Initial momentum $P_i = 0$.
Final momentum $P_f = m_{boat} v_b + m_{man} v_m = 0$.
Substituting the values: $200 v_b + 50(v_b + 2) = 0$.
$200 v_b + 50 v_b + 100 = 0$.
$250 v_b = -100$.
$v_b = -100 / 250 = -0.4 \,ms^{-1}$.
The magnitude of the velocity is $0.4 \,ms^{-1}$ or $2/5 \,ms^{-1}$.
Given, the velocity of the man with respect to the boat is $v_{m/b} = 2 \,ms^{-1}$.
By relative velocity definition, $v_{m/b} = v_m - v_b$, so $v_m = v_b + 2$.
Since there is no external horizontal force on the system (man + boat), the momentum of the system is conserved.
Initial momentum $P_i = 0$.
Final momentum $P_f = m_{boat} v_b + m_{man} v_m = 0$.
Substituting the values: $200 v_b + 50(v_b + 2) = 0$.
$200 v_b + 50 v_b + 100 = 0$.
$250 v_b = -100$.
$v_b = -100 / 250 = -0.4 \,ms^{-1}$.
The magnitude of the velocity is $0.4 \,ms^{-1}$ or $2/5 \,ms^{-1}$.
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8
PhysicsDifficultMCQTS EAMCET · 2006
To the free end of a spring hanging from a rigid support,a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly,the amount of maximum stretching is: (force constant,$k$)
A
$\frac{m g}{k}$
B
$2 d$
C
$\frac{m g}{3 k}$
D
$4 d$
Solution
(B) Case $1$: When the block is lowered slowly,it reaches equilibrium where the spring force balances the gravitational force. $k d = m g$,so $d = \frac{m g}{k}$.
Case $2$: When the block is allowed to fall suddenly from the unstretched position,the block undergoes simple harmonic motion. Let the maximum extension be $x$. By the principle of conservation of energy,the loss in gravitational potential energy equals the gain in elastic potential energy of the spring.
$m g x = \frac{1}{2} k x^2$.
Solving for $x$,we get $x = \frac{2 m g}{k}$.
Since $d = \frac{m g}{k}$,we have $x = 2 d$.
Case $2$: When the block is allowed to fall suddenly from the unstretched position,the block undergoes simple harmonic motion. Let the maximum extension be $x$. By the principle of conservation of energy,the loss in gravitational potential energy equals the gain in elastic potential energy of the spring.
$m g x = \frac{1}{2} k x^2$.
Solving for $x$,we get $x = \frac{2 m g}{k}$.
Since $d = \frac{m g}{k}$,we have $x = 2 d$.
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9
PhysicsDifficultMCQTS EAMCET · 2006
An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected,the coefficient of viscosity of the liquid is approximately,(in $Pa \ s$):
A
$13000$
B
$1300$
C
$130$
D
$13$
Solution
(C) The terminal velocity $v$ of an air bubble rising in a liquid is given by the formula: $v = \frac{2}{9} \frac{r^2 \rho g}{\eta}$.
Here,$r = 1 \ cm = 10^{-2} \ m$,$\rho = 1.5 \ g/cc = 1.5 \times 10^3 \ kg/m^3$,$g = 9.8 \ m/s^2$,and $v = 0.25 \ cm/s = 0.25 \times 10^{-2} \ m/s$.
Rearranging the formula for the coefficient of viscosity $\eta$: $\eta = \frac{2}{9} \cdot \frac{r^2 \rho g}{v}$.
Substituting the values: $\eta = \frac{2}{9} \cdot \frac{(10^{-2})^2 \cdot (1.5 \times 10^3) \cdot 9.8}{0.25 \times 10^{-2}}$.
$\eta = \frac{2}{9} \cdot \frac{10^{-4} \cdot 1500 \cdot 9.8}{0.0025} = \frac{2}{9} \cdot \frac{1.47}{0.0025} = \frac{2}{9} \cdot 588 \approx 130.6 \ Pa \ s$.
Thus,the coefficient of viscosity is approximately $130 \ Pa \ s$.
Here,$r = 1 \ cm = 10^{-2} \ m$,$\rho = 1.5 \ g/cc = 1.5 \times 10^3 \ kg/m^3$,$g = 9.8 \ m/s^2$,and $v = 0.25 \ cm/s = 0.25 \times 10^{-2} \ m/s$.
Rearranging the formula for the coefficient of viscosity $\eta$: $\eta = \frac{2}{9} \cdot \frac{r^2 \rho g}{v}$.
Substituting the values: $\eta = \frac{2}{9} \cdot \frac{(10^{-2})^2 \cdot (1.5 \times 10^3) \cdot 9.8}{0.25 \times 10^{-2}}$.
$\eta = \frac{2}{9} \cdot \frac{10^{-4} \cdot 1500 \cdot 9.8}{0.0025} = \frac{2}{9} \cdot \frac{1.47}{0.0025} = \frac{2}{9} \cdot 588 \approx 130.6 \ Pa \ s$.
Thus,the coefficient of viscosity is approximately $130 \ Pa \ s$.
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10
PhysicsDifficultMCQTS EAMCET · 2006
Two blocks of masses $1 \, kg$ and $2 \, kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $2 \times 10^9 \, N/m^2$. What should be the minimum radius of the wire used if it is not to break? Take $g = 10 \, m/s^2$.
A
$4.6 \times 10^{-5} \, m$
B
$4.6 \times 10^{-6} \, m$
C
$2.5 \times 10^{-6} \, m$
D
$2.5 \times 10^{-5} \, m$
Solution
(A) The stress in the wire is given by $\text{Stress} = \frac{\text{Tension}}{\text{Area of cross-section}}$.
To avoid breaking, the stress must not exceed the breaking stress.
Let the tension in the wire be $T$ and the acceleration of the system be $a$.
The equations of motion for the two blocks are:
For the $1 \, kg$ block: $T - 1(10) = 1a \implies T - 10 = a$ (Equation $1$)
For the $2 \, kg$ block: $2(10) - T = 2a \implies 20 - T = 2a$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(T - 10) + (20 - T) = a + 2a$
$10 = 3a \implies a = \frac{10}{3} \, m/s^2$
Substituting $a$ into Equation $1$:
$T = 10 + \frac{10}{3} = \frac{40}{3} \, N$
The breaking stress is $\sigma_{max} = 2 \times 10^9 \, N/m^2$. The area of cross-section is $A = \pi r^2$.
Setting the stress equal to the breaking stress to find the minimum radius $r$:
$\frac{T}{A} = \sigma_{max} \implies \frac{40/3}{\pi r^2} = 2 \times 10^9$
$r^2 = \frac{40}{3 \times \pi \times 2 \times 10^9} = \frac{20}{3 \pi \times 10^9} \approx \frac{20}{9.4247 \times 10^9} \approx 2.122 \times 10^{-9} \, m^2$
$r = \sqrt{2.122 \times 10^{-9}} \approx 4.6 \times 10^{-5} \, m$.
To avoid breaking, the stress must not exceed the breaking stress.
Let the tension in the wire be $T$ and the acceleration of the system be $a$.
The equations of motion for the two blocks are:
For the $1 \, kg$ block: $T - 1(10) = 1a \implies T - 10 = a$ (Equation $1$)
For the $2 \, kg$ block: $2(10) - T = 2a \implies 20 - T = 2a$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(T - 10) + (20 - T) = a + 2a$
$10 = 3a \implies a = \frac{10}{3} \, m/s^2$
Substituting $a$ into Equation $1$:
$T = 10 + \frac{10}{3} = \frac{40}{3} \, N$
The breaking stress is $\sigma_{max} = 2 \times 10^9 \, N/m^2$. The area of cross-section is $A = \pi r^2$.
Setting the stress equal to the breaking stress to find the minimum radius $r$:
$\frac{T}{A} = \sigma_{max} \implies \frac{40/3}{\pi r^2} = 2 \times 10^9$
$r^2 = \frac{40}{3 \times \pi \times 2 \times 10^9} = \frac{20}{3 \pi \times 10^9} \approx \frac{20}{9.4247 \times 10^9} \approx 2.122 \times 10^{-9} \, m^2$
$r = \sqrt{2.122 \times 10^{-9}} \approx 4.6 \times 10^{-5} \, m$.
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11
PhysicsDifficultMCQTS EAMCET · 2006
$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$,the maximum height reached by it is: (in meters)
A
$5 \sqrt{3}$
B
$\frac{5}{\sqrt{3}}$
C
$\frac{10}{\sqrt{3}}$
D
$10 \sqrt{3}$
Solution
(B) The formula for the range of a projectile is $R = \frac{u^2 \sin 2\theta}{g}$.
Given $R = 20 \ m$ and $\theta = 30^{\circ}$,we have $20 = \frac{u^2 \sin(2 \times 30^{\circ})}{g}$.
$\Rightarrow 20 = \frac{u^2 \sin 60^{\circ}}{g} = \frac{u^2}{g} \times \frac{\sqrt{3}}{2}$.
Therefore,$\frac{u^2}{g} = \frac{20 \times 2}{\sqrt{3}} = \frac{40}{\sqrt{3}}$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left(\frac{u^2}{g}\right) \sin^2 \theta$.
Substituting the values,$H = \frac{1}{2} \times \left(\frac{40}{\sqrt{3}}\right) \times \sin^2 30^{\circ}$.
$H = \frac{20}{\sqrt{3}} \times \left(\frac{1}{2}\right)^2 = \frac{20}{\sqrt{3}} \times \frac{1}{4} = \frac{5}{\sqrt{3}} \ m$.
Given $R = 20 \ m$ and $\theta = 30^{\circ}$,we have $20 = \frac{u^2 \sin(2 \times 30^{\circ})}{g}$.
$\Rightarrow 20 = \frac{u^2 \sin 60^{\circ}}{g} = \frac{u^2}{g} \times \frac{\sqrt{3}}{2}$.
Therefore,$\frac{u^2}{g} = \frac{20 \times 2}{\sqrt{3}} = \frac{40}{\sqrt{3}}$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left(\frac{u^2}{g}\right) \sin^2 \theta$.
Substituting the values,$H = \frac{1}{2} \times \left(\frac{40}{\sqrt{3}}\right) \times \sin^2 30^{\circ}$.
$H = \frac{20}{\sqrt{3}} \times \left(\frac{1}{2}\right)^2 = \frac{20}{\sqrt{3}} \times \frac{1}{4} = \frac{5}{\sqrt{3}} \ m$.
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12
PhysicsDifficultMCQTS EAMCET · 2006
$A$ man standing on a road has to hold his umbrella at $30^{\circ}$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \,km/h$. He finds that raindrops are hitting his head vertically. The actual speed of raindrops is:
A
$20 \,km/h$
B
$10 \sqrt{3} \,km/h$
C
$20 \sqrt{3} \,km/h$
D
$10 \,km/h$
Solution
$(\text{A})$ Let $\vec{v}_{r,g}$ be the velocity of rain with respect to the ground, $\vec{v}_{m,g}$ be the velocity of the man with respect to the ground, and $\vec{v}_{r,m}$ be the velocity of rain with respect to the man.
When the man is at rest, the rain falls at $30^{\circ}$ with the vertical. Thus, the horizontal component of $\vec{v}_{r,g}$ is $v_{r,g} \sin 30^{\circ}$ and the vertical component is $v_{r,g} \cos 30^{\circ}$.
When the man runs at $v_{m,g} = 10 \,km/h$, the rain appears to fall vertically. This means the horizontal component of the relative velocity $\vec{v}_{r,m} = \vec{v}_{r,g} - \vec{v}_{m,g}$ must be zero.
Therefore, the horizontal component of $\vec{v}_{r,g}$ must be equal to the velocity of the man:
$v_{r,g} \sin 30^{\circ} = v_{m,g}$
$v_{r,g} \times (1/2) = 10 \,km/h$
$v_{r,g} = 20 \,km/h$.
When the man is at rest, the rain falls at $30^{\circ}$ with the vertical. Thus, the horizontal component of $\vec{v}_{r,g}$ is $v_{r,g} \sin 30^{\circ}$ and the vertical component is $v_{r,g} \cos 30^{\circ}$.
When the man runs at $v_{m,g} = 10 \,km/h$, the rain appears to fall vertically. This means the horizontal component of the relative velocity $\vec{v}_{r,m} = \vec{v}_{r,g} - \vec{v}_{m,g}$ must be zero.
Therefore, the horizontal component of $\vec{v}_{r,g}$ must be equal to the velocity of the man:
$v_{r,g} \sin 30^{\circ} = v_{m,g}$
$v_{r,g} \times (1/2) = 10 \,km/h$
$v_{r,g} = 20 \,km/h$.
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13
PhysicsMediumMCQTS EAMCET · 2006
The bob of a simple pendulum is hanging vertically down from a fixed identical bob by means of a string of length $l$. If both bobs are charged with a charge $q$ each,what is the time period of the pendulum? (Ignore the radii of the bobs.)
A
$2 \pi \sqrt{\frac{l}{g+\frac{q^2}{l^2 m}}}$
B
$2 \pi \sqrt{\frac{l}{g-\frac{q^2}{l^2 m}}}$
C
$2 \pi \sqrt{\frac{l}{g}}$
D
$2 \pi \sqrt{\frac{l}{g-\frac{q^2}{l}}}$
Solution
(C) The bob of the simple pendulum is hanging vertically below a fixed bob. Both bobs carry an identical charge $q$.
Since the bobs are aligned vertically,the electrostatic force of repulsion acts along the line joining the two bobs,which is the same line as the string.
This electrostatic force acts in the upward direction on the moving bob,opposing the gravitational force (weight $mg$).
However,for a simple pendulum,the restoring force is provided by the component of gravity perpendicular to the string,which is $mg \sin \theta$.
The electrostatic force acts along the string and does not have a component perpendicular to the string.
Therefore,the electrostatic force does not affect the restoring force or the motion of the pendulum.
Thus,the time period of the pendulum remains unchanged: $T = 2 \pi \sqrt{\frac{l}{g}}$.
Since the bobs are aligned vertically,the electrostatic force of repulsion acts along the line joining the two bobs,which is the same line as the string.
This electrostatic force acts in the upward direction on the moving bob,opposing the gravitational force (weight $mg$).
However,for a simple pendulum,the restoring force is provided by the component of gravity perpendicular to the string,which is $mg \sin \theta$.
The electrostatic force acts along the string and does not have a component perpendicular to the string.
Therefore,the electrostatic force does not affect the restoring force or the motion of the pendulum.
Thus,the time period of the pendulum remains unchanged: $T = 2 \pi \sqrt{\frac{l}{g}}$.
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14
PhysicsMediumMCQTS EAMCET · 2006
For a system to follow the law of conservation of linear momentum during a collision,the condition is:
$(1)$ Total external force acting on the system is zero.
$(2)$ Total external force acting on the system is finite and the time of collision is negligible.
$(3)$ Total internal force acting on the system is zero.
$(1)$ Total external force acting on the system is zero.
$(2)$ Total external force acting on the system is finite and the time of collision is negligible.
$(3)$ Total internal force acting on the system is zero.
A
$(1)$ only
B
$(2)$ only
C
$(3)$ only
D
$(1)$ or $(2)$
Solution
(D) According to Newton's second law of motion,the rate of change of linear momentum is equal to the net external force acting on the system:
$F_{ext} = \frac{dp}{dt}$
If the total external force $F_{ext} = 0$,then $\frac{dp}{dt} = 0$,which implies that the linear momentum $p$ is constant (conserved).
In the case of a collision,if the time interval $\Delta t$ is extremely small (negligible) and the external force is finite,the impulse $J = F_{ext} \cdot \Delta t$ approaches zero.
Therefore,for a collision,the linear momentum is conserved if the external force is zero $OR$ if the external force is finite and the collision time is negligible.
Thus,both conditions $(1)$ and $(2)$ satisfy the requirement for the conservation of linear momentum.
$F_{ext} = \frac{dp}{dt}$
If the total external force $F_{ext} = 0$,then $\frac{dp}{dt} = 0$,which implies that the linear momentum $p$ is constant (conserved).
In the case of a collision,if the time interval $\Delta t$ is extremely small (negligible) and the external force is finite,the impulse $J = F_{ext} \cdot \Delta t$ approaches zero.
Therefore,for a collision,the linear momentum is conserved if the external force is zero $OR$ if the external force is finite and the collision time is negligible.
Thus,both conditions $(1)$ and $(2)$ satisfy the requirement for the conservation of linear momentum.
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15
PhysicsDifficultMCQTS EAMCET · 2006
$A$ uniform rod of length $8 a$ and mass $6 m$ lies on a smooth horizontal surface. Two point masses $m$ and $2 m$ moving in the same plane with speed $2 v$ and $v$ respectively strike the rod perpendicularly at distances $a$ and $2 a$ from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is:
A
$\frac{6 v}{32 a}$
B
$\frac{6 v}{33 a}$
C
$\frac{6 v}{40 a}$
D
$\frac{6 v}{41 a}$
Solution
(D) Applying the principle of conservation of angular momentum about the center of mass (midpoint $O$) of the rod:
Initial angular momentum $L_i = m(2v)(a) + (2m)(v)(2a) = 2mav + 4mav = 6mav$.
Final angular momentum $L_f = I_{total} \omega$,where $I_{total}$ is the moment of inertia of the system about the center of mass after the particles stick to the rod.
$I_{total} = I_{rod} + I_{m} + I_{2m} = \frac{(6m)(8a)^2}{12} + m(a)^2 + (2m)(2a)^2$.
$I_{total} = \frac{6m(64a^2)}{12} + ma^2 + 8ma^2 = 32ma^2 + ma^2 + 8ma^2 = 41ma^2$.
Equating $L_i = L_f$:
$6mav = (41ma^2) \omega$.
Therefore,$\omega = \frac{6mav}{41ma^2} = \frac{6v}{41a}$.
Initial angular momentum $L_i = m(2v)(a) + (2m)(v)(2a) = 2mav + 4mav = 6mav$.
Final angular momentum $L_f = I_{total} \omega$,where $I_{total}$ is the moment of inertia of the system about the center of mass after the particles stick to the rod.
$I_{total} = I_{rod} + I_{m} + I_{2m} = \frac{(6m)(8a)^2}{12} + m(a)^2 + (2m)(2a)^2$.
$I_{total} = \frac{6m(64a^2)}{12} + ma^2 + 8ma^2 = 32ma^2 + ma^2 + 8ma^2 = 41ma^2$.
Equating $L_i = L_f$:
$6mav = (41ma^2) \omega$.
Therefore,$\omega = \frac{6mav}{41ma^2} = \frac{6v}{41a}$.
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16
PhysicsMediumMCQTS EAMCET · 2006
The temperature of a thin uniform circular disc of $1 \ m$ diameter is increased by $10^{\circ} C$. The percentage increase in the moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face is: (linear coefficient of expansion $\alpha = 11 \times 10^{-6} /{ }^{\circ} C$)
A
$0.0055$
B
$0.011$
C
$0.022$
D
$0.044$
Solution
(C) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.
Since the mass $M$ remains constant,the change in moment of inertia depends on the change in radius $R$.
For a small change in temperature $\Delta t$,the change in radius is given by $\Delta R = R \alpha \Delta t$.
The new radius is $R' = R(1 + \alpha \Delta t)$.
The new moment of inertia is $I' = \frac{1}{2} M (R')^2 = \frac{1}{2} M R^2 (1 + \alpha \Delta t)^2$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$,we get $I' \approx \frac{1}{2} M R^2 (1 + 2 \alpha \Delta t) = I(1 + 2 \alpha \Delta t)$.
The fractional change in moment of inertia is $\frac{\Delta I}{I} = \frac{I' - I}{I} = 2 \alpha \Delta t$.
Substituting the given values: $\alpha = 11 \times 10^{-6} /{ }^{\circ} C$ and $\Delta t = 10^{\circ} C$.
$\frac{\Delta I}{I} = 2 \times (11 \times 10^{-6}) \times 10 = 220 \times 10^{-6} = 0.00022$.
The percentage increase is $0.00022 \times 100 = 0.022 \%$.
Since the mass $M$ remains constant,the change in moment of inertia depends on the change in radius $R$.
For a small change in temperature $\Delta t$,the change in radius is given by $\Delta R = R \alpha \Delta t$.
The new radius is $R' = R(1 + \alpha \Delta t)$.
The new moment of inertia is $I' = \frac{1}{2} M (R')^2 = \frac{1}{2} M R^2 (1 + \alpha \Delta t)^2$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$,we get $I' \approx \frac{1}{2} M R^2 (1 + 2 \alpha \Delta t) = I(1 + 2 \alpha \Delta t)$.
The fractional change in moment of inertia is $\frac{\Delta I}{I} = \frac{I' - I}{I} = 2 \alpha \Delta t$.
Substituting the given values: $\alpha = 11 \times 10^{-6} /{ }^{\circ} C$ and $\Delta t = 10^{\circ} C$.
$\frac{\Delta I}{I} = 2 \times (11 \times 10^{-6}) \times 10 = 220 \times 10^{-6} = 0.00022$.
The percentage increase is $0.00022 \times 100 = 0.022 \%$.
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17
PhysicsDifficultMCQTS EAMCET · 2006
Assume the earth's orbit around the sun as circular and the distance between their centres as $D$. The mass of the earth is $M$ and its radius is $R$. If the earth has an angular velocity $\omega_0$ with respect to its centre and $\omega$ with respect to the centre of the sun,the total kinetic energy of the earth is:
A
$\frac{M R^2 \omega_0^2}{5}\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
B
$\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
C
$\frac{2}{5} M R^2 \omega_0^2\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
D
$\frac{2}{5} M R^2 \omega_0^2\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]$
Solution
(B) The total kinetic energy of the earth is the sum of its rotational kinetic energy about its own axis and its translational kinetic energy due to its orbital motion around the sun.
Rotational kinetic energy $K_{rot} = \frac{1}{2} I \omega_0^2$,where $I = \frac{2}{5} M R^2$ for a solid sphere.
So,$K_{rot} = \frac{1}{2} \times \frac{2}{5} M R^2 \omega_0^2 = \frac{1}{5} M R^2 \omega_0^2$.
Translational kinetic energy $K_{trans} = \frac{1}{2} M v^2$,where $v = D \omega$.
So,$K_{trans} = \frac{1}{2} M (D \omega)^2 = \frac{1}{2} M D^2 \omega^2$.
Total kinetic energy $K_{total} = K_{rot} + K_{trans} = \frac{1}{5} M R^2 \omega_0^2 + \frac{1}{2} M D^2 \omega^2$.
Factoring out $\frac{M R^2 \omega_0^2}{5}$,we get $K_{total} = \frac{M R^2 \omega_0^2}{5} \left[ 1 + \frac{5}{2} \frac{M D^2 \omega^2}{M R^2 \omega_0^2} \right] = \frac{M R^2 \omega_0^2}{5} \left[ 1 + \frac{5}{2} \left( \frac{D \omega}{R \omega_0} \right)^2 \right]$.
Rotational kinetic energy $K_{rot} = \frac{1}{2} I \omega_0^2$,where $I = \frac{2}{5} M R^2$ for a solid sphere.
So,$K_{rot} = \frac{1}{2} \times \frac{2}{5} M R^2 \omega_0^2 = \frac{1}{5} M R^2 \omega_0^2$.
Translational kinetic energy $K_{trans} = \frac{1}{2} M v^2$,where $v = D \omega$.
So,$K_{trans} = \frac{1}{2} M (D \omega)^2 = \frac{1}{2} M D^2 \omega^2$.
Total kinetic energy $K_{total} = K_{rot} + K_{trans} = \frac{1}{5} M R^2 \omega_0^2 + \frac{1}{2} M D^2 \omega^2$.
Factoring out $\frac{M R^2 \omega_0^2}{5}$,we get $K_{total} = \frac{M R^2 \omega_0^2}{5} \left[ 1 + \frac{5}{2} \frac{M D^2 \omega^2}{M R^2 \omega_0^2} \right] = \frac{M R^2 \omega_0^2}{5} \left[ 1 + \frac{5}{2} \left( \frac{D \omega}{R \omega_0} \right)^2 \right]$.
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18
PhysicsDifficultMCQTS EAMCET · 2006
Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :
A
$r_A / r_B$
B
$r_B / r_A$
C
$r_A^2 / r_B^2$
D
$r_B^2 / r_A^2$
Solution
(B) According to the Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.
Since $Q = mc\Delta T = (\rho V c) \Delta T$,the rate of change of temperature is $\frac{dT}{dt} = \frac{1}{mc} \frac{dQ}{dt}$.
For a sphere,$V = \frac{4}{3} \pi r^3$ and $A = 4 \pi r^2$.
Substituting these,we get $\frac{dT}{dt} = \frac{\sigma (4 \pi r^2) (T^4 - T_0^4)}{\rho (\frac{4}{3} \pi r^3) c} = \frac{3 \sigma (T^4 - T_0^4)}{\rho r c}$.
Thus,the rate of change of temperature is inversely proportional to the radius: $\frac{dT}{dt} \propto \frac{1}{r}$.
Therefore,the ratio of the rate of change of temperature for spheres $A$ and $B$ is $\frac{(dT/dt)_A}{(dT/dt)_B} = \frac{r_B}{r_A}$.
Since $Q = mc\Delta T = (\rho V c) \Delta T$,the rate of change of temperature is $\frac{dT}{dt} = \frac{1}{mc} \frac{dQ}{dt}$.
For a sphere,$V = \frac{4}{3} \pi r^3$ and $A = 4 \pi r^2$.
Substituting these,we get $\frac{dT}{dt} = \frac{\sigma (4 \pi r^2) (T^4 - T_0^4)}{\rho (\frac{4}{3} \pi r^3) c} = \frac{3 \sigma (T^4 - T_0^4)}{\rho r c}$.
Thus,the rate of change of temperature is inversely proportional to the radius: $\frac{dT}{dt} \propto \frac{1}{r}$.
Therefore,the ratio of the rate of change of temperature for spheres $A$ and $B$ is $\frac{(dT/dt)_A}{(dT/dt)_B} = \frac{r_B}{r_A}$.
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19
PhysicsDifficultMCQTS EAMCET · 2006
$A$ cyclic process $ABCD$ is shown in the given $P-V$ diagram. Which of the following diagrams represents the same process in a $P-T$ graph?
A
B
C
D
Solution
(A) In the $P-V$ diagram:
$1$. Process $AB$ is an isobaric process,where pressure $P$ remains constant. In a $P-T$ diagram,this is represented by a horizontal line.
$2$. Process $BC$ is an isothermal process,where temperature $T$ remains constant. In a $P-T$ diagram,this is represented by a vertical line.
$3$. Process $CD$ is an isochoric process,where volume $V$ remains constant. Since $PV = nRT$,for constant $V$,$P \propto T$. Thus,in a $P-T$ diagram,this is a straight line passing through the origin.
$4$. Process $DA$ is an adiabatic process,which is represented by a curve in the $P-T$ diagram.
Comparing these characteristics,the correct $P-T$ diagram is shown in option $(a)$.
$1$. Process $AB$ is an isobaric process,where pressure $P$ remains constant. In a $P-T$ diagram,this is represented by a horizontal line.
$2$. Process $BC$ is an isothermal process,where temperature $T$ remains constant. In a $P-T$ diagram,this is represented by a vertical line.
$3$. Process $CD$ is an isochoric process,where volume $V$ remains constant. Since $PV = nRT$,for constant $V$,$P \propto T$. Thus,in a $P-T$ diagram,this is a straight line passing through the origin.
$4$. Process $DA$ is an adiabatic process,which is represented by a curve in the $P-T$ diagram.
Comparing these characteristics,the correct $P-T$ diagram is shown in option $(a)$.
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20
PhysicsDifficultMCQTS EAMCET · 2006
$A$ given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be $0.75$ of its initial pressure. The ratio of the specific heats of the gas is approximately:
A
$1.2$
B
$1.41$
C
$1.67$
D
$1.83$
Solution
(B) In an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P \propto \frac{1}{V}$.
Let the initial pressure be $P$ and initial volume be $V_1$. After isothermal compression,pressure $P_2 = 2P$ and volume $V_2 = V_1/2$.
Thus,$\frac{V_1}{V_2} = 2$.
Now,the gas expands adiabatically to its original volume $V_1$. Let the final pressure be $P_3 = 0.75P$.
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_1^\gamma$.
Substituting the values: $(2P) \left(\frac{V_1}{2}\right)^\gamma = (0.75P) V_1^\gamma$.
$2 \cdot \left(\frac{1}{2}\right)^\gamma = 0.75$.
$2^{1-\gamma} = \frac{3}{4} = 3 \cdot 2^{-2}$.
$2^{3-\gamma} = 3$.
Taking log on both sides: $(3-\gamma) \log 2 = \log 3$.
$3-\gamma = \frac{\log 3}{\log 2} \approx 1.585$.
$\gamma = 3 - 1.585 = 1.415 \approx 1.41$.
Let the initial pressure be $P$ and initial volume be $V_1$. After isothermal compression,pressure $P_2 = 2P$ and volume $V_2 = V_1/2$.
Thus,$\frac{V_1}{V_2} = 2$.
Now,the gas expands adiabatically to its original volume $V_1$. Let the final pressure be $P_3 = 0.75P$.
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_1^\gamma$.
Substituting the values: $(2P) \left(\frac{V_1}{2}\right)^\gamma = (0.75P) V_1^\gamma$.
$2 \cdot \left(\frac{1}{2}\right)^\gamma = 0.75$.
$2^{1-\gamma} = \frac{3}{4} = 3 \cdot 2^{-2}$.
$2^{3-\gamma} = 3$.
Taking log on both sides: $(3-\gamma) \log 2 = \log 3$.
$3-\gamma = \frac{\log 3}{\log 2} \approx 1.585$.
$\gamma = 3 - 1.585 = 1.415 \approx 1.41$.
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21
PhysicsDifficultMCQTS EAMCET · 2006
An observer is standing $500 \,m$ away from a vertical hill. Starting between the observer and the hill, a police van sounding a siren of frequency $1000 \,Hz$ moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is $970 \,Hz$, the frequency of the sound heard after reflection from the hill (in $Hz$) is about, (velocity of sound $= 330 \,m/s$):
A
$1042$
B
$1032$
C
$1022$
D
$1012$
Solution
(B) The frequency of the sound heard directly by the observer moving away from the source is given by the Doppler effect formula: $f_1 = f_0 \left( \frac{v}{v + v_s} \right)$.
Given $f_1 = 970 \,Hz$, $f_0 = 1000 \,Hz$, and $v = 330 \,m/s$, we have: $970 = 1000 \left( \frac{330}{330 + v_s} \right)$.
Solving for $v_s$: $330 + v_s = \frac{1000 \times 330}{970} \approx 340.2 \,m/s$, so $v_s \approx 10.2 \,m/s$.
The sound reflected from the hill acts as if it is coming from a source moving towards the observer. The frequency of the reflected sound $f_2$ is given by: $f_2 = f_0 \left( \frac{v}{v - v_s} \right)$.
Substituting the values: $f_2 = 1000 \left( \frac{330}{330 - 10.2} \right) = 1000 \left( \frac{330}{319.8} \right) \approx 1031.89 \,Hz$.
Rounding to the nearest integer, the frequency is approximately $1032 \,Hz$.
Given $f_1 = 970 \,Hz$, $f_0 = 1000 \,Hz$, and $v = 330 \,m/s$, we have: $970 = 1000 \left( \frac{330}{330 + v_s} \right)$.
Solving for $v_s$: $330 + v_s = \frac{1000 \times 330}{970} \approx 340.2 \,m/s$, so $v_s \approx 10.2 \,m/s$.
The sound reflected from the hill acts as if it is coming from a source moving towards the observer. The frequency of the reflected sound $f_2$ is given by: $f_2 = f_0 \left( \frac{v}{v - v_s} \right)$.
Substituting the values: $f_2 = 1000 \left( \frac{330}{330 - 10.2} \right) = 1000 \left( \frac{330}{319.8} \right) \approx 1031.89 \,Hz$.
Rounding to the nearest integer, the frequency is approximately $1032 \,Hz$.
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22
PhysicsDifficultMCQTS EAMCET · 2006
Two strings $A$ and $B$ of lengths $L_A = 80 \text{ cm}$ and $L_B = x \text{ cm}$ respectively are used separately in a sonometer. The ratio of their densities $(d_A / d_B)$ is $0.81$. The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency,the value of $x$ is:
A
$33$
B
$102$
C
$144$
D
$130$
Solution
(C) Given: $T_A = T_B$,$f_A = f_B$,$L_A = 80 \text{ cm}$,$L_B = x \text{ cm}$.
Ratio of densities: $\frac{d_A}{d_B} = 0.81$.
Ratio of diameters: $\frac{D_A}{D_B} = 2$.
Linear mass density $\mu = \text{Area} \times \text{density} = \frac{\pi D^2}{4} \times d$.
Therefore,$\frac{\mu_A}{\mu_B} = \left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B} = (2)^2 \times 0.81 = 4 \times 0.81 = 3.24$.
The fundamental frequency of a string is $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $f_A = f_B$ and $T_A = T_B$,we have $L_A \sqrt{\mu_A} = L_B \sqrt{\mu_B}$.
$\frac{L_B}{L_A} = \sqrt{\frac{\mu_A}{\mu_B}} = \sqrt{3.24} = 1.8$.
$x = 80 \times 1.8 = 144 \text{ cm}$.
Ratio of densities: $\frac{d_A}{d_B} = 0.81$.
Ratio of diameters: $\frac{D_A}{D_B} = 2$.
Linear mass density $\mu = \text{Area} \times \text{density} = \frac{\pi D^2}{4} \times d$.
Therefore,$\frac{\mu_A}{\mu_B} = \left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B} = (2)^2 \times 0.81 = 4 \times 0.81 = 3.24$.
The fundamental frequency of a string is $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $f_A = f_B$ and $T_A = T_B$,we have $L_A \sqrt{\mu_A} = L_B \sqrt{\mu_B}$.
$\frac{L_B}{L_A} = \sqrt{\frac{\mu_A}{\mu_B}} = \sqrt{3.24} = 1.8$.
$x = 80 \times 1.8 = 144 \text{ cm}$.
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23
PhysicsDifficultMCQTS EAMCET · 2006
$A$ motor is used to deliver water at a certain rate through a given horizontal pipe. To deliver $n$-times the water through the same pipe in the same time,the power of the motor must be increased as follows:
A
$n$-times
B
$n^2$-times
C
$n^3$-times
D
$n^4$-times
Solution
(C) Let the density of water be $\rho$,the cross-sectional area of the pipe be $A$,and the velocity of water be $v$.
The mass of water delivered per second is given by $m = A v \rho$ ...$(i)$.
The power $P$ required to deliver this water is equal to the rate of change of kinetic energy of the water:
$P = \frac{1}{2} m v^2 = \frac{1}{2} (A v \rho) v^2 = \frac{1}{2} A \rho v^3$ ...(ii).
If we want to deliver $n$-times the mass of water in the same time,the new mass flow rate $m' = n m$.
Since $m = A v \rho$,we have $n (A v \rho) = A v' \rho$,which implies $v' = n v$.
The new power $P'$ required is $P' = \frac{1}{2} A \rho (v')^3$.
Taking the ratio of the new power to the original power:
$\frac{P'}{P} = \frac{\frac{1}{2} A \rho (n v)^3}{\frac{1}{2} A \rho v^3} = n^3$.
Therefore,the power must be increased $n^3$-times.
The mass of water delivered per second is given by $m = A v \rho$ ...$(i)$.
The power $P$ required to deliver this water is equal to the rate of change of kinetic energy of the water:
$P = \frac{1}{2} m v^2 = \frac{1}{2} (A v \rho) v^2 = \frac{1}{2} A \rho v^3$ ...(ii).
If we want to deliver $n$-times the mass of water in the same time,the new mass flow rate $m' = n m$.
Since $m = A v \rho$,we have $n (A v \rho) = A v' \rho$,which implies $v' = n v$.
The new power $P'$ required is $P' = \frac{1}{2} A \rho (v')^3$.
Taking the ratio of the new power to the original power:
$\frac{P'}{P} = \frac{\frac{1}{2} A \rho (n v)^3}{\frac{1}{2} A \rho v^3} = n^3$.
Therefore,the power must be increased $n^3$-times.
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24
PhysicsDifficultMCQTS EAMCET · 2006
$A$ bullet of mass $10 \,g$ is fired horizontally with a velocity $1000 \,ms^{-1}$ from a rifle situated at a height $50 \,m$ above the ground. If the bullet reaches the ground with a velocity $500 \,ms^{-1}$, the work done against air resistance in the trajectory of the bullet is : $(g=10 \,ms^{-2})$ (in $\,J$)
A
$5005$
B
$3755$
C
$3750$
D
$17.5$
Solution
(B) According to the work-energy theorem, the work done by all forces equals the change in kinetic energy.
$W_{gravity} + W_{air} = \Delta K$
$mgh + W_{air} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$
Given: $m = 10 \,g = 0.01 \,kg$, $h = 50 \,m$, $v_i = 1000 \,ms^{-1}$, $v_f = 500 \,ms^{-1}$, $g = 10 \,ms^{-2}$.
$W_{gravity} = mgh = 0.01 \times 10 \times 50 = 5 \,J$.
Change in kinetic energy $\Delta K = \frac{1}{2} \times 0.01 \times (500^2 - 1000^2) = 0.005 \times (250000 - 1000000) = 0.005 \times (-750000) = -3750 \,J$.
Substituting these values into the work-energy theorem:
$5 + W_{air} = -3750$
$W_{air} = -3750 - 5 = -3755 \,J$.
The work done $AGAINST$ air resistance is the negative of the work done $BY$ air resistance.
Work done against air resistance $= -(-3755 \,J) = 3755 \,J$.
$W_{gravity} + W_{air} = \Delta K$
$mgh + W_{air} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$
Given: $m = 10 \,g = 0.01 \,kg$, $h = 50 \,m$, $v_i = 1000 \,ms^{-1}$, $v_f = 500 \,ms^{-1}$, $g = 10 \,ms^{-2}$.
$W_{gravity} = mgh = 0.01 \times 10 \times 50 = 5 \,J$.
Change in kinetic energy $\Delta K = \frac{1}{2} \times 0.01 \times (500^2 - 1000^2) = 0.005 \times (250000 - 1000000) = 0.005 \times (-750000) = -3750 \,J$.
Substituting these values into the work-energy theorem:
$5 + W_{air} = -3750$
$W_{air} = -3750 - 5 = -3755 \,J$.
The work done $AGAINST$ air resistance is the negative of the work done $BY$ air resistance.
Work done against air resistance $= -(-3755 \,J) = 3755 \,J$.
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25
PhysicsEasyMCQTS EAMCET · 2006
$A$ free neutron decays spontaneously into :
A
a proton,an electron and anti-neutrino
B
a proton,an electron and a-neutrino
C
a proton and electron
D
a proton,and electron,a neutrino and an anti-neutrino.
Solution
(A) free neutron is unstable and decays spontaneously into a proton,an electron,and an electron anti-neutrino.
The decay equation is given by:
$n \rightarrow p + e^{-} + \bar{\nu}_{e}$
This process is known as beta-minus $(\beta^{-})$ decay,which conserves charge,baryon number,and lepton number.
The decay equation is given by:
$n \rightarrow p + e^{-} + \bar{\nu}_{e}$
This process is known as beta-minus $(\beta^{-})$ decay,which conserves charge,baryon number,and lepton number.
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26
PhysicsMediumMCQTS EAMCET · 2006
Match the appropriate pairs from List-$I$ and List-$II$:
| List-$I$ | List-$II$ |
|---|---|
| $1$. Nitrogen molecules | $(A)$ Continuous spectrum |
| $2$. Incandescent solids | $(B)$ Absorption spectrum |
| $3$. Fraunhofer lines | $(C)$ Band spectrum |
| $4$. Electric arc between iron rods | $(D)$ Emission spectrum |
A
$1-C, 2-A, 3-B, 4-D$
B
$1-B, 2-A, 3-D, 4-C$
C
$1-D, 2-A, 3-B, 4-C$
D
$1-A, 2-C, 3-D, 4-B$
Solution
(A) $1$. Nitrogen molecules exhibit a band spectrum because molecular spectra consist of bands due to transitions between electronic,vibrational,and rotational energy levels.
$2$. Incandescent solids emit a continuous spectrum because they contain a wide range of frequencies due to thermal agitation.
$3$. Fraunhofer lines are dark lines observed in the solar spectrum,which are caused by the absorption of specific wavelengths by gases in the solar atmosphere,thus forming an absorption spectrum.
$4$. An electric arc between iron rods produces a line emission spectrum characteristic of the iron atoms present in the arc.
Therefore,the correct matching is $1-C, 2-A, 3-B, 4-D$.
$2$. Incandescent solids emit a continuous spectrum because they contain a wide range of frequencies due to thermal agitation.
$3$. Fraunhofer lines are dark lines observed in the solar spectrum,which are caused by the absorption of specific wavelengths by gases in the solar atmosphere,thus forming an absorption spectrum.
$4$. An electric arc between iron rods produces a line emission spectrum characteristic of the iron atoms present in the arc.
Therefore,the correct matching is $1-C, 2-A, 3-B, 4-D$.
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27
PhysicsDifficultMCQTS EAMCET · 2006
An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)
A
$5$
B
$10$
C
$15$
D
$20$
Solution
(D) Initially,the oil drop is in equilibrium,so the electric force balances the gravitational force: $qE = mg$.
Here,$m = 4.8 \times 10^{-10} \,g = 4.8 \times 10^{-13} \,kg$,$q = 2.4 \times 10^{-18} \,C$,and $g = 10 \,m/s^2$.
When the polarity of the plates is changed,the direction of the electric force $qE$ reverses and acts downwards along with the gravitational force $mg$.
The new net force on the drop is $F_{net} = qE + mg$.
Since $qE = mg$,we have $F_{net} = mg + mg = 2mg$.
The instantaneous acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{2mg}{m} = 2g$.
Substituting $g = 10 \,m/s^2$,we get $a = 2 \times 10 = 20 \,m/s^2$.
Here,$m = 4.8 \times 10^{-10} \,g = 4.8 \times 10^{-13} \,kg$,$q = 2.4 \times 10^{-18} \,C$,and $g = 10 \,m/s^2$.
When the polarity of the plates is changed,the direction of the electric force $qE$ reverses and acts downwards along with the gravitational force $mg$.
The new net force on the drop is $F_{net} = qE + mg$.
Since $qE = mg$,we have $F_{net} = mg + mg = 2mg$.
The instantaneous acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{2mg}{m} = 2g$.
Substituting $g = 10 \,m/s^2$,we get $a = 2 \times 10 = 20 \,m/s^2$.
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28
PhysicsDifficultMCQTS EAMCET · 2006
Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(D) Let the number of wrongly connected cells in the $12$-cell battery be $m$.
Each wrongly connected cell cancels the emf of one correctly connected cell.
Thus,the effective emf of the $12$-cell battery is $E_{12} = (12 - m)E - mE = (12 - 2m)E$.
When the $12$-cell battery and the $2$-cell battery (emf $2E$) aid each other,the total emf is $E_{total} = (12 - 2m)E + 2E = (14 - 2m)E$.
The current is $i_1 = \frac{(14 - 2m)E}{R} = 3 \text{ A}$ ...$(i)$
When they oppose each other,the total emf is $E_{total} = (12 - 2m)E - 2E = (10 - 2m)E$.
The current is $i_2 = \frac{(10 - 2m)E}{R} = 2 \text{ A}$ ...(ii)
Dividing equation $(i)$ by (ii):
$\frac{3}{2} = \frac{14 - 2m}{10 - 2m}$
$3(10 - 2m) = 2(14 - 2m)$
$30 - 6m = 28 - 4m$
$2 = 2m$
$m = 1$.
Therefore,the number of wrongly connected cells is $1$.
Each wrongly connected cell cancels the emf of one correctly connected cell.
Thus,the effective emf of the $12$-cell battery is $E_{12} = (12 - m)E - mE = (12 - 2m)E$.
When the $12$-cell battery and the $2$-cell battery (emf $2E$) aid each other,the total emf is $E_{total} = (12 - 2m)E + 2E = (14 - 2m)E$.
The current is $i_1 = \frac{(14 - 2m)E}{R} = 3 \text{ A}$ ...$(i)$
When they oppose each other,the total emf is $E_{total} = (12 - 2m)E - 2E = (10 - 2m)E$.
The current is $i_2 = \frac{(10 - 2m)E}{R} = 2 \text{ A}$ ...(ii)
Dividing equation $(i)$ by (ii):
$\frac{3}{2} = \frac{14 - 2m}{10 - 2m}$
$3(10 - 2m) = 2(14 - 2m)$
$30 - 6m = 28 - 4m$
$2 = 2m$
$m = 1$.
Therefore,the number of wrongly connected cells is $1$.
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29
PhysicsDifficultMCQTS EAMCET · 2006
One end each of a resistance $r$,a capacitor $C$,and a resistance $2r$ are connected together. The other ends are respectively connected to the positive terminals of batteries $P$,$Q$,and $R$ having emfs $E$,$E$,and $2E$. The negative terminals of the batteries are then connected together. In this circuit,with steady current,the potential drop across the capacitor is:
A
$\frac{E}{3}$
B
$\frac{E}{2}$
C
$\frac{2E}{3}$
D
$E$
Solution
(A) In the steady state,no current flows through the capacitor branch.
Let the common potential of the junction where the resistor $r$,capacitor $C$,and resistor $2r$ meet be $V_x$,and the potential of the junction where the negative terminals of the batteries meet be $0 \text{ V}$.
The potential at the positive terminals of the batteries $P$,$Q$,and $R$ are $E$,$E$,and $2E$ respectively.
Since no current flows through the capacitor branch,the potential at the capacitor plate connected to the junction is $V_x$,and the potential at the other plate is $E$.
Applying Kirchhoff's Current Law at the junction $V_x$:
$\frac{V_x - E}{r} + \frac{V_x - 2E}{2r} + 0 = 0$
Multiplying by $2r$:
$2(V_x - E) + (V_x - 2E) = 0$
$2V_x - 2E + V_x - 2E = 0$
$3V_x = 4E$
$V_x = \frac{4E}{3}$
The potential drop across the capacitor is $|V_x - E| = |\frac{4E}{3} - E| = \frac{E}{3}$.
Let the common potential of the junction where the resistor $r$,capacitor $C$,and resistor $2r$ meet be $V_x$,and the potential of the junction where the negative terminals of the batteries meet be $0 \text{ V}$.
The potential at the positive terminals of the batteries $P$,$Q$,and $R$ are $E$,$E$,and $2E$ respectively.
Since no current flows through the capacitor branch,the potential at the capacitor plate connected to the junction is $V_x$,and the potential at the other plate is $E$.
Applying Kirchhoff's Current Law at the junction $V_x$:
$\frac{V_x - E}{r} + \frac{V_x - 2E}{2r} + 0 = 0$
Multiplying by $2r$:
$2(V_x - E) + (V_x - 2E) = 0$
$2V_x - 2E + V_x - 2E = 0$
$3V_x = 4E$
$V_x = \frac{4E}{3}$
The potential drop across the capacitor is $|V_x - E| = |\frac{4E}{3} - E| = \frac{E}{3}$.
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30
PhysicsDifficultMCQTS EAMCET · 2006
$A$ rectangular loop of length $l$ and breadth $b$ is placed at a distance of $x$ from an infinitely long wire carrying current $i$ such that the direction of the current is parallel to the breadth of the loop. If the loop moves away from the current-carrying wire in a direction perpendicular to it with a velocity $v$,the magnitude of the induced emf in the loop is: ($\mu_0=$ permeability of free space)
A
$\frac{\mu_0 i v}{2 \pi x}\left(\frac{l+b}{b}\right)$
B
$\frac{\mu_0 i^2 v}{4 \pi^2 x} \log \left(\frac{b}{l}\right)$
C
$\frac{\mu_0 i l b v}{2 \pi x(l+x)}$
D
$\frac{\mu_0 i l b v}{2 \pi} \log \left(\frac{x+l}{x}\right)$
Solution
(C) The magnetic field $B$ at a distance $r$ from an infinitely long wire is given by $B = \frac{\mu_0 i}{2 \pi r}$.
As the loop moves with velocity $v$,the motional emf is induced in the two vertical sides of length $b$.
For the side at distance $x$,the velocity is perpendicular to the magnetic field,so the induced emf is $e_1 = B_1 v b = \left(\frac{\mu_0 i}{2 \pi x}\right) v b$.
For the side at distance $(x+l)$,the induced emf is $e_2 = B_2 v b = \left(\frac{\mu_0 i}{2 \pi (x+l)}\right) v b$.
Since the loop is moving away,the emf induced in these two sides opposes each other in the loop circuit.
The net magnitude of the emf is $e = |e_1 - e_2| = \frac{\mu_0 i v b}{2 \pi} \left( \frac{1}{x} - \frac{1}{x+l} \right)$.
Simplifying the expression: $e = \frac{\mu_0 i v b}{2 \pi} \left( \frac{x+l-x}{x(x+l)} \right) = \frac{\mu_0 i l b v}{2 \pi x(x+l)}$.
As the loop moves with velocity $v$,the motional emf is induced in the two vertical sides of length $b$.
For the side at distance $x$,the velocity is perpendicular to the magnetic field,so the induced emf is $e_1 = B_1 v b = \left(\frac{\mu_0 i}{2 \pi x}\right) v b$.
For the side at distance $(x+l)$,the induced emf is $e_2 = B_2 v b = \left(\frac{\mu_0 i}{2 \pi (x+l)}\right) v b$.
Since the loop is moving away,the emf induced in these two sides opposes each other in the loop circuit.
The net magnitude of the emf is $e = |e_1 - e_2| = \frac{\mu_0 i v b}{2 \pi} \left( \frac{1}{x} - \frac{1}{x+l} \right)$.
Simplifying the expression: $e = \frac{\mu_0 i v b}{2 \pi} \left( \frac{x+l-x}{x(x+l)} \right) = \frac{\mu_0 i l b v}{2 \pi x(x+l)}$.
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31
PhysicsDifficultMCQTS EAMCET · 2006
$A$ small square loop of wire of side $l$ is placed inside a large square loop of side $L$ $(L \gg l)$. If the loops are coplanar and their centres coincide,the mutual inductance of the system is directly proportional to :
A
$\frac{L}{l}$
B
$\frac{l}{L}$
C
$\frac{L^2}{l}$
D
$\frac{l^2}{L}$
Solution
(D) Let a current $I$ flow through the large square loop of side $L$. The magnetic field $B$ produced by one side of the large loop at its centre is given by the formula for a finite wire: $B_{side} = \frac{\mu_0 I}{4 \pi d} (\sin \alpha + \sin \beta)$,where $d = L/2$ and $\alpha = \beta = 45^\circ$.
Since there are four such sides,the total magnetic field at the centre is $B = 4 \times \frac{\mu_0 I}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{\pi L} \times 2 \times \frac{2}{\sqrt{2}} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.
Since $L \gg l$,we can assume the magnetic field $B$ is uniform over the area of the smaller loop $S_2 = l^2$.
The magnetic flux linked with the smaller loop is $\phi_2 = B \times S_2 = \left( \frac{2 \sqrt{2} \mu_0 I}{\pi L} \right) l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I} = \frac{2 \sqrt{2} \mu_0 l^2}{\pi L}$.
Therefore,$M \propto \frac{l^2}{L}$.
Since there are four such sides,the total magnetic field at the centre is $B = 4 \times \frac{\mu_0 I}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{\pi L} \times 2 \times \frac{2}{\sqrt{2}} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.
Since $L \gg l$,we can assume the magnetic field $B$ is uniform over the area of the smaller loop $S_2 = l^2$.
The magnetic flux linked with the smaller loop is $\phi_2 = B \times S_2 = \left( \frac{2 \sqrt{2} \mu_0 I}{\pi L} \right) l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I} = \frac{2 \sqrt{2} \mu_0 l^2}{\pi L}$.
Therefore,$M \propto \frac{l^2}{L}$.
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32
PhysicsDifficultMCQTS EAMCET · 2006
Along the $x$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at a point $P$ located at a distance $r$ from the charge $-q$ $(r > a)$ is ($\varepsilon_0$ is the permittivity of free space):
A
$\frac{q a}{4 \pi \varepsilon_0 r^2}$
B
$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
C
$\frac{q(a^2/4)}{4 \pi \varepsilon_0 r^3}$
D
$\frac{q}{4 \pi \varepsilon_0 r}$
Solution
(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$. Point $P$ is at distance $r$ from the charge $-q$ at $x=a$. Since $P$ is on the $x$-axis at distance $r$ from $x=a$,its coordinate is $x = a + r$.
The potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
Factoring out $q/2$:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$
Combining the terms:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$
Simplifying the numerator:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right] = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$
Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$
The potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
Factoring out $q/2$:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$
Combining the terms:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$
Simplifying the numerator:
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right] = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$
Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$
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33
PhysicsDifficultMCQTS EAMCET · 2006
Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a << r$) is: ($\varepsilon_0$ is the permittivity of free space)
A
$\frac{q a}{4 \pi \varepsilon_0 r^2}$
B
$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
C
$\frac{q(a^2/4)}{4 \pi \varepsilon_0 r^3}$
D
$\frac{q}{4 \pi \varepsilon_0 r}$
Solution
(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
Let the point $P$ be at $x = a + r$. The distances of the charges from $P$ are:
For charge $\frac{q}{2}$ at $x=0$: distance $d_1 = (a+r) - 0 = r+a$
For charge $-q$ at $x=a$: distance $d_2 = (a+r) - a = r$
For charge $\frac{q}{2}$ at $x=2a$: distance $d_3 = |(a+r) - 2a| = |r-a| = r-a$ (since $r >> a$)
The total potential $V_P$ is:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$
Since $r >> a$,we have $r^2 - a^2 \approx r^2$.
Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.
Let the point $P$ be at $x = a + r$. The distances of the charges from $P$ are:
For charge $\frac{q}{2}$ at $x=0$: distance $d_1 = (a+r) - 0 = r+a$
For charge $-q$ at $x=a$: distance $d_2 = (a+r) - a = r$
For charge $\frac{q}{2}$ at $x=2a$: distance $d_3 = |(a+r) - 2a| = |r-a| = r-a$ (since $r >> a$)
The total potential $V_P$ is:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$
Since $r >> a$,we have $r^2 - a^2 \approx r^2$.
Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.
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34
PhysicsDifficultMCQTS EAMCET · 2006
Two short magnets $AB$ and $CD$ are in the $X-Y$ plane and are parallel to the $X$-axis. The coordinates of their centres are $(0,2)$ and $(2,0)$ respectively. The line joining the north-south poles of $CD$ is opposite to that of $AB$ and lies along the positive $X$-axis. The resultant magnetic field induction due to $AB$ and $CD$ at a point $P(2,2)$ is $100 \times 10^{-7} \ T$. When the poles of the magnet $CD$ are reversed,the resultant field induction is $50 \times 10^{-7} \ T$. The values of the magnetic moments of $AB$ and $CD$ (in $Am^2$) are:
A
$300; 200$
B
$600; 400$
C
$200; 100$
D
$300; 150$
Solution
(A) Let $M_1$ be the magnetic moment of magnet $AB$ and $M_2$ be the magnetic moment of magnet $CD$. Point $P(2,2)$ lies on the axial line of magnet $AB$ at a distance $r_1 = 2$ from its centre,and on the equatorial line of magnet $CD$ at a distance $r_2 = 2$ from its centre.
The magnetic field due to $AB$ at $P$ is $B_1 = \frac{\mu_0}{4\pi} \frac{2M_1}{r_1^3} = 10^{-7} \times \frac{2M_1}{2^3} = 10^{-7} \times \frac{M_1}{4}$.
The magnetic field due to $CD$ at $P$ is $B_2 = \frac{\mu_0}{4\pi} \frac{M_2}{r_2^3} = 10^{-7} \times \frac{M_2}{2^3} = 10^{-7} \times \frac{M_2}{8}$.
Given the resultant field is $100 \times 10^{-7} \ T$,we have $B_1 + B_2 = 100 \times 10^{-7}$.
$10^{-7} (\frac{M_1}{4} + \frac{M_2}{8}) = 100 \times 10^{-7} \Rightarrow 2M_1 + M_2 = 800$ $(i)$.
When the poles of $CD$ are reversed,the field $B_2$ changes direction,so $B_1 - B_2 = 50 \times 10^{-7}$.
$10^{-7} (\frac{M_1}{4} - \frac{M_2}{8}) = 50 \times 10^{-7} \Rightarrow 2M_1 - M_2 = 400$ (ii).
Adding $(i)$ and (ii): $4M_1 = 1200 \Rightarrow M_1 = 300 \ Am^2$.
Substituting $M_1$ in $(i)$: $2(300) + M_2 = 800 \Rightarrow M_2 = 200 \ Am^2$.
The magnetic field due to $AB$ at $P$ is $B_1 = \frac{\mu_0}{4\pi} \frac{2M_1}{r_1^3} = 10^{-7} \times \frac{2M_1}{2^3} = 10^{-7} \times \frac{M_1}{4}$.
The magnetic field due to $CD$ at $P$ is $B_2 = \frac{\mu_0}{4\pi} \frac{M_2}{r_2^3} = 10^{-7} \times \frac{M_2}{2^3} = 10^{-7} \times \frac{M_2}{8}$.
Given the resultant field is $100 \times 10^{-7} \ T$,we have $B_1 + B_2 = 100 \times 10^{-7}$.
$10^{-7} (\frac{M_1}{4} + \frac{M_2}{8}) = 100 \times 10^{-7} \Rightarrow 2M_1 + M_2 = 800$ $(i)$.
When the poles of $CD$ are reversed,the field $B_2$ changes direction,so $B_1 - B_2 = 50 \times 10^{-7}$.
$10^{-7} (\frac{M_1}{4} - \frac{M_2}{8}) = 50 \times 10^{-7} \Rightarrow 2M_1 - M_2 = 400$ (ii).
Adding $(i)$ and (ii): $4M_1 = 1200 \Rightarrow M_1 = 300 \ Am^2$.
Substituting $M_1$ in $(i)$: $2(300) + M_2 = 800 \Rightarrow M_2 = 200 \ Am^2$.
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35
PhysicsDifficultMCQTS EAMCET · 2006
$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:
A
$1: 2: 4$
B
$1: \sqrt{2}: 1$
C
$2: \sqrt{2}: 1$
D
$1: 1: 2$
Solution
(B) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
Since kinetic energy $K$ and magnetic field $B$ are the same for all particles, we have $r \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: mass $m_p = m$, charge $q_p = q$. So, $r_p \propto \frac{\sqrt{m}}{q}$.
For a deuteron $(d)$: mass $m_d = 2m$, charge $q_d = q$. So, $r_d \propto \frac{\sqrt{2m}}{q}$.
For an $\alpha$-particle $(\alpha)$: mass $m_\alpha = 4m$, charge $q_\alpha = 2q$. So, $r_\alpha \propto \frac{\sqrt{4m}}{2q} = \frac{2\sqrt{m}}{2q} = \frac{\sqrt{m}}{q}$.
Therefore, the ratio $r_p : r_d : r_\alpha = \frac{\sqrt{m}}{q} : \frac{\sqrt{2m}}{q} : \frac{\sqrt{m}}{q} = 1 : \sqrt{2} : 1$.
Since kinetic energy $K$ and magnetic field $B$ are the same for all particles, we have $r \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: mass $m_p = m$, charge $q_p = q$. So, $r_p \propto \frac{\sqrt{m}}{q}$.
For a deuteron $(d)$: mass $m_d = 2m$, charge $q_d = q$. So, $r_d \propto \frac{\sqrt{2m}}{q}$.
For an $\alpha$-particle $(\alpha)$: mass $m_\alpha = 4m$, charge $q_\alpha = 2q$. So, $r_\alpha \propto \frac{\sqrt{4m}}{2q} = \frac{2\sqrt{m}}{2q} = \frac{\sqrt{m}}{q}$.
Therefore, the ratio $r_p : r_d : r_\alpha = \frac{\sqrt{m}}{q} : \frac{\sqrt{2m}}{q} : \frac{\sqrt{m}}{q} = 1 : \sqrt{2} : 1$.
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36
PhysicsMediumMCQTS EAMCET · 2006
When a positively charged particle enters a uniform magnetic field with uniform velocity,its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix
A
$(1)$ only
B
$(1)$ or $(2)$
C
$(1)$ or $(3)$
D
any one of $(1)$,$(2)$ and $(3)$
Solution
(D) The force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Case $(1)$: If the velocity $\vec{v}$ is parallel or anti-parallel to the magnetic field $\vec{B}$,then $\vec{v} \times \vec{B} = 0$. The force is zero,and the particle continues in a straight line.
Case $(2)$: If the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the magnetic force acts as a centripetal force,causing the particle to move in a circle.
Case $(3)$: If the velocity $\vec{v}$ makes an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$) with the magnetic field $\vec{B}$,the velocity can be resolved into two components: one parallel to $\vec{B}$ (causing linear motion) and one perpendicular to $\vec{B}$ (causing circular motion). The resultant trajectory is a helix.
Therefore,all three trajectories are possible depending on the angle between the velocity and the magnetic field.
Case $(1)$: If the velocity $\vec{v}$ is parallel or anti-parallel to the magnetic field $\vec{B}$,then $\vec{v} \times \vec{B} = 0$. The force is zero,and the particle continues in a straight line.
Case $(2)$: If the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the magnetic force acts as a centripetal force,causing the particle to move in a circle.
Case $(3)$: If the velocity $\vec{v}$ makes an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$) with the magnetic field $\vec{B}$,the velocity can be resolved into two components: one parallel to $\vec{B}$ (causing linear motion) and one perpendicular to $\vec{B}$ (causing circular motion). The resultant trajectory is a helix.
Therefore,all three trajectories are possible depending on the angle between the velocity and the magnetic field.
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37
PhysicsMediumMCQTS EAMCET · 2006
The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:
A
both torque and net force are present
B
torque is present but no net force
C
both torque and net force are absent
D
net force is present but no torque
Solution
(B) When a magnetic needle is placed in a uniform magnetic field,the magnetic field exerts a force $F = mB$ on the north pole and $F = -mB$ on the south pole,where $m$ is the pole strength and $B$ is the magnetic field intensity.
Since the forces are equal in magnitude and opposite in direction,the net force on the needle is $F_{net} = mB - mB = 0$.
However,because these forces act at different points (the poles),they form a couple that exerts a torque $\tau = mB \times l \sin(\theta)$ on the needle,which tends to align it with the magnetic field.
Therefore,a torque is present,but the net force is zero.
Since the forces are equal in magnitude and opposite in direction,the net force on the needle is $F_{net} = mB - mB = 0$.
However,because these forces act at different points (the poles),they form a couple that exerts a torque $\tau = mB \times l \sin(\theta)$ on the needle,which tends to align it with the magnetic field.
Therefore,a torque is present,but the net force is zero.
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38
PhysicsEasyMCQTS EAMCET · 2006
$A$ free neutron decays spontaneously into:
A
a proton, an electron and anti-neutrino
B
a proton, an electron and a-neutrino
C
a proton and electron
D
a proton, an electron, a neutrino and an anti-neutrino
Solution
(A) free neutron is unstable and decays spontaneously into a proton, an electron, and an electron anti-neutrino. The decay process is represented by the equation: $n \rightarrow p + e^{-} + \bar{\nu}_{e}$. This process is known as beta-minus $(\beta^{-})$ decay.
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39
PhysicsDifficultMCQTS EAMCET · 2006
The two surfaces of a biconvex lens have the same radii of curvature. This lens is made of glass of refractive index $1.5$ and has a focal length $10 \ cm$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index $= 4/3$), its focal length (in $cm$) is:
A
$5$
B
$10$
C
$20$
D
$40$
Solution
(D) For a biconvex lens with radii of curvature $R$ and refractive index $\mu_g = 1.5$, the lens maker's formula in air is:
$\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (1.5 - 1) \left( \frac{2}{R} \right) = \frac{1}{R}$
Given $f = 10 \ cm$, we have $R = 10 \ cm$.
When the lens is cut perpendicular to the principal axis, we get two plano-convex lenses, each with one flat surface $(R_1 = \infty)$ and one curved surface $(R_2 = -10 \ cm)$.
The focal length $f'$ of each plano-convex lens is given by:
$\frac{1}{f'} = (\mu_g - 1) \left( \frac{1}{\infty} - \frac{1}{-10} \right) = 0.5 \times \frac{1}{10} = \frac{1}{20} \implies f' = 20 \ cm$.
When these two pieces are glued such that the convex surfaces touch each other, the combination acts as a single lens with two curved surfaces of radii $R_1 = 10 \ cm$ and $R_2 = -10 \ cm$ (biconvex again).
When immersed in water $(\mu_w = 4/3)$, the new focal length $F'$ is given by:
$\frac{1}{F'} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$\frac{1}{F'} = \left( \frac{1.5}{4/3} - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right) = \left( \frac{4.5}{4} - 1 \right) \left( \frac{2}{10} \right) = \left( \frac{0.5}{4} \right) \left( \frac{1}{5} \right) = \frac{1}{8} \times \frac{1}{5} = \frac{1}{40}$
Therefore, $F' = 40 \ cm$.
$\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (1.5 - 1) \left( \frac{2}{R} \right) = \frac{1}{R}$
Given $f = 10 \ cm$, we have $R = 10 \ cm$.
When the lens is cut perpendicular to the principal axis, we get two plano-convex lenses, each with one flat surface $(R_1 = \infty)$ and one curved surface $(R_2 = -10 \ cm)$.
The focal length $f'$ of each plano-convex lens is given by:
$\frac{1}{f'} = (\mu_g - 1) \left( \frac{1}{\infty} - \frac{1}{-10} \right) = 0.5 \times \frac{1}{10} = \frac{1}{20} \implies f' = 20 \ cm$.
When these two pieces are glued such that the convex surfaces touch each other, the combination acts as a single lens with two curved surfaces of radii $R_1 = 10 \ cm$ and $R_2 = -10 \ cm$ (biconvex again).
When immersed in water $(\mu_w = 4/3)$, the new focal length $F'$ is given by:
$\frac{1}{F'} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$\frac{1}{F'} = \left( \frac{1.5}{4/3} - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right) = \left( \frac{4.5}{4} - 1 \right) \left( \frac{2}{10} \right) = \left( \frac{0.5}{4} \right) \left( \frac{1}{5} \right) = \frac{1}{8} \times \frac{1}{5} = \frac{1}{40}$
Therefore, $F' = 40 \ cm$.
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40
PhysicsEasyMCQTS EAMCET · 2006
Dispersive power depends on the following:
A
material of the prism
B
shape of the prism
C
size of the prism
D
size,shape and material of the prism
Solution
(A) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion $(\delta_v - \delta_r)$ to the mean deviation $(\delta_y)$.
Mathematically, $\omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{(\mu_v - 1)A - (\mu_r - 1)A}{(\mu_y - 1)A} = \frac{\mu_v - \mu_r}{\mu_y - 1}$.
From this formula, it is clear that the dispersive power depends only on the refractive indices $(\mu_v, \mu_r, \mu_y)$ of the material of the prism for different colors.
It is independent of the angle of the prism $(A)$, the shape, or the size of the prism.
Therefore, the correct option is $A$.
Mathematically, $\omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{(\mu_v - 1)A - (\mu_r - 1)A}{(\mu_y - 1)A} = \frac{\mu_v - \mu_r}{\mu_y - 1}$.
From this formula, it is clear that the dispersive power depends only on the refractive indices $(\mu_v, \mu_r, \mu_y)$ of the material of the prism for different colors.
It is independent of the angle of the prism $(A)$, the shape, or the size of the prism.
Therefore, the correct option is $A$.
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41
PhysicsDifficultMCQTS EAMCET · 2006
Consider a $p-n$ junction as a capacitor,formed with $p$ and $n$-materials acting as thin metal electrodes and depletion layer width acting as separation between them. Basing on this,assume that an $n-p-n$ transistor is working as an amplifier in $CE$ configuration. If $C_1$ and $C_2$ are the base-emitter and collector-emitter junction capacitances,then :
A
$C_1 > C_2$
B
$C_1 < C_2$
C
$C_1 = C_2$
D
$C_1 = C_2 = 0$
Solution
(A) In an $n-p-n$ transistor working in $CE$ configuration,the base-emitter junction is forward-biased,while the collector-emitter junction is reverse-biased.
The capacitance of a junction is given by $C = \frac{\epsilon A}{d}$,where $d$ is the width of the depletion layer.
For a forward-biased junction (base-emitter),the depletion layer width $d_1$ is very small.
For a reverse-biased junction (collector-emitter),the depletion layer width $d_2$ is significantly larger.
Since $C \propto \frac{1}{d}$,a smaller depletion width results in a larger capacitance.
Therefore,$d_1 < d_2$ implies $C_1 > C_2$.
The capacitance of a junction is given by $C = \frac{\epsilon A}{d}$,where $d$ is the width of the depletion layer.
For a forward-biased junction (base-emitter),the depletion layer width $d_1$ is very small.
For a reverse-biased junction (collector-emitter),the depletion layer width $d_2$ is significantly larger.
Since $C \propto \frac{1}{d}$,a smaller depletion width results in a larger capacitance.
Therefore,$d_1 < d_2$ implies $C_1 > C_2$.
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42
PhysicsDifficultMCQTS EAMCET · 2006
If the cold junction is held at $0^{\circ} C$,the thermo emf $V$ of a thermocouple varies as $V = 10 \times 10^{-6} t - \frac{1}{40} \times 10^{-6} t^2$,where $t$ is the temperature of the hot junction in $^{\circ} C$. The neutral temperature and the maximum value of thermo emf are respectively:
A
$200^{\circ} C ; 2 \text{ mV}$
B
$400^{\circ} C ; 2 \text{ mV}$
C
$100^{\circ} C ; 1 \text{ mV}$
D
$200^{\circ} C ; 1 \text{ mV}$
Solution
(D) Given the thermo emf equation: $V = 10 \times 10^{-6} t - \frac{1}{40} \times 10^{-6} t^2$.
To find the neutral temperature $(t_n)$,we differentiate $V$ with respect to $t$ and set it to zero: $\frac{dV}{dt} = 10 \times 10^{-6} - \frac{2}{40} \times 10^{-6} t = 0$.
$10 \times 10^{-6} = \frac{1}{20} \times 10^{-6} t_n$.
$t_n = 200^{\circ} C$.
To find the maximum thermo emf $(V_{\max})$,we substitute $t_n = 200^{\circ} C$ into the original equation:
$V_{\max} = 10 \times 10^{-6} (200) - \frac{1}{40} \times 10^{-6} (200)^2$.
$V_{\max} = 2 \times 10^{-3} - \frac{40000}{40} \times 10^{-6} = 2 \times 10^{-3} - 1 \times 10^{-3} = 1 \times 10^{-3} \text{ V} = 1 \text{ mV}$.
To find the neutral temperature $(t_n)$,we differentiate $V$ with respect to $t$ and set it to zero: $\frac{dV}{dt} = 10 \times 10^{-6} - \frac{2}{40} \times 10^{-6} t = 0$.
$10 \times 10^{-6} = \frac{1}{20} \times 10^{-6} t_n$.
$t_n = 200^{\circ} C$.
To find the maximum thermo emf $(V_{\max})$,we substitute $t_n = 200^{\circ} C$ into the original equation:
$V_{\max} = 10 \times 10^{-6} (200) - \frac{1}{40} \times 10^{-6} (200)^2$.
$V_{\max} = 2 \times 10^{-3} - \frac{40000}{40} \times 10^{-6} = 2 \times 10^{-3} - 1 \times 10^{-3} = 1 \times 10^{-3} \text{ V} = 1 \text{ mV}$.
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43
PhysicsMediumMCQTS EAMCET · 2006
If $C, R, L$ and $I$ denote capacity,resistance,inductance and electric current respectively,the quantities having the same dimensions of time are :
$(1)$ $C R$
$(2)$ $\frac{L}{R}$
$(3)$ $\sqrt{L C}$
$(4)$ $L I^2$
$(1)$ $C R$
$(2)$ $\frac{L}{R}$
$(3)$ $\sqrt{L C}$
$(4)$ $L I^2$
A
$(1)$ and $(2)$ only
B
$(1)$ and $(3)$ only
C
$(1)$ and $(4)$ only
D
$(1)$,$(2)$ and $(3)$ only
Solution
(D) The dimensional formulas are:
$[C] = [M^{-1} L^{-2} T^4 A^2]$
$[R] = [M L^2 T^{-3} A^{-2}]$
$[L] = [M L^2 T^{-2} A^{-2}]$
$[I] = [A]$
$(1)$ $[CR] = [M^{-1} L^{-2} T^4 A^2] \times [M L^2 T^{-3} A^{-2}] = [T^1]$. This represents time.
$(2)$ $[L/R] = [M L^2 T^{-2} A^{-2}] / [M L^2 T^{-3} A^{-2}] = [T^1]$. This represents time.
$(3)$ $[\sqrt{LC}] = ([M L^2 T^{-2} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2])^{1/2} = [T^2]^{1/2} = [T^1]$. This represents time.
$(4)$ $[LI^2] = [M L^2 T^{-2} A^{-2}] \times [A^2] = [M L^2 T^{-2}]$. This represents energy,not time.
Thus,quantities $(1)$,$(2)$,and $(3)$ have the dimensions of time. Therefore,option $(d)$ is correct.
$[C] = [M^{-1} L^{-2} T^4 A^2]$
$[R] = [M L^2 T^{-3} A^{-2}]$
$[L] = [M L^2 T^{-2} A^{-2}]$
$[I] = [A]$
$(1)$ $[CR] = [M^{-1} L^{-2} T^4 A^2] \times [M L^2 T^{-3} A^{-2}] = [T^1]$. This represents time.
$(2)$ $[L/R] = [M L^2 T^{-2} A^{-2}] / [M L^2 T^{-3} A^{-2}] = [T^1]$. This represents time.
$(3)$ $[\sqrt{LC}] = ([M L^2 T^{-2} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2])^{1/2} = [T^2]^{1/2} = [T^1]$. This represents time.
$(4)$ $[LI^2] = [M L^2 T^{-2} A^{-2}] \times [A^2] = [M L^2 T^{-2}]$. This represents energy,not time.
Thus,quantities $(1)$,$(2)$,and $(3)$ have the dimensions of time. Therefore,option $(d)$ is correct.
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44
PhysicsDifficultMCQTS EAMCET · 2006
In Young's double slit experiment,the first slit has a width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is: (in $: 1$)
A
$2$
B
$4$
C
$9$
D
$8$
Solution
(C) The intensity of light $I$ is directly proportional to the width $w$ of the slit,i.e.,$I \propto w$.
Given that the width of the first slit $w_1 = 4w_2$,the intensities are related as $I_1 = 4I_2$.
Let $I_2 = I$,then $I_1 = 4I$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{4I} + \sqrt{I})^2}{(\sqrt{4I} - \sqrt{I})^2} = \frac{(2\sqrt{I} + \sqrt{I})^2}{(2\sqrt{I} - \sqrt{I})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(3\sqrt{I})^2}{(\sqrt{I})^2} = \frac{9I}{I} = \frac{9}{1}$
Thus,the ratio is $9: 1$.
Given that the width of the first slit $w_1 = 4w_2$,the intensities are related as $I_1 = 4I_2$.
Let $I_2 = I$,then $I_1 = 4I$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{4I} + \sqrt{I})^2}{(\sqrt{4I} - \sqrt{I})^2} = \frac{(2\sqrt{I} + \sqrt{I})^2}{(2\sqrt{I} - \sqrt{I})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(3\sqrt{I})^2}{(\sqrt{I})^2} = \frac{9I}{I} = \frac{9}{1}$
Thus,the ratio is $9: 1$.
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