PhysicsQ1–8 of 8 questions
Page 1 of 1 · English
1
PhysicsEasyMCQAIIMS · 1985
Which of the following is not a unit of energy?
A
$W \cdot s$
B
$kg \cdot m/s$
C
$N \cdot m$
D
Joule
Solution
(B) Energy is defined as the capacity to do work. The $SI$ unit of energy is the Joule $(J)$.
$1 \text{ Joule} = 1 \text{ Newton} \cdot \text{ meter} (N \cdot m) = 1 \text{ Watt} \cdot \text{ second} (W \cdot s)$.
Option $(b)$ $kg \cdot m/s$ represents the unit of linear momentum $(p = mv)$,not energy.
Therefore,the correct option is $(b)$.
$1 \text{ Joule} = 1 \text{ Newton} \cdot \text{ meter} (N \cdot m) = 1 \text{ Watt} \cdot \text{ second} (W \cdot s)$.
Option $(b)$ $kg \cdot m/s$ represents the unit of linear momentum $(p = mv)$,not energy.
Therefore,the correct option is $(b)$.
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2
PhysicsMediumMCQAIIMS · 1985
The dimension of Planck's constant is equal to that of:
A
Energy
B
Linear momentum
C
Power
D
Angular momentum
Solution
(D) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Thus,the dimension of $h$ is $[h] = [E] / [\nu]$.
$[E] = [M L^2 T^{-2}]$ and $[\nu] = [T^{-1}]$.
So,$[h] = [M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$.
Angular momentum $L$ is defined as $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is radius.
$[L] = [M] [L T^{-1}] [L] = [M L^2 T^{-1}]$.
Therefore,the dimension of Planck's constant is equal to that of angular momentum.
Thus,the dimension of $h$ is $[h] = [E] / [\nu]$.
$[E] = [M L^2 T^{-2}]$ and $[\nu] = [T^{-1}]$.
So,$[h] = [M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$.
Angular momentum $L$ is defined as $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is radius.
$[L] = [M] [L T^{-1}] [L] = [M L^2 T^{-1}]$.
Therefore,the dimension of Planck's constant is equal to that of angular momentum.
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3
PhysicsEasyMCQAIIMS · 1985
$A$ mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity $\omega_0$. If the length of the string and angular velocity are doubled,the tension in the string which was initially $T_0$ is now
A
$T_0$
B
$T_0/2$
C
$4T_0$
D
$8T_0$
Solution
(D) The tension in the string provides the necessary centripetal force for circular motion.
Initially,the tension $T_0$ is given by $T_0 = mR\omega_0^2$,where $m$ is the mass,$R$ is the initial length of the string,and $\omega_0$ is the initial angular velocity.
In the second case,the new length $R' = 2R$ and the new angular velocity $\omega' = 2\omega_0$.
The new tension $T$ is given by $T = mR'(\omega')^2$.
Substituting the new values: $T = m(2R)(2\omega_0)^2 = m(2R)(4\omega_0^2) = 8mR\omega_0^2$.
Since $T_0 = mR\omega_0^2$,we have $T = 8T_0$.
Initially,the tension $T_0$ is given by $T_0 = mR\omega_0^2$,where $m$ is the mass,$R$ is the initial length of the string,and $\omega_0$ is the initial angular velocity.
In the second case,the new length $R' = 2R$ and the new angular velocity $\omega' = 2\omega_0$.
The new tension $T$ is given by $T = mR'(\omega')^2$.
Substituting the new values: $T = m(2R)(2\omega_0)^2 = m(2R)(4\omega_0^2) = 8mR\omega_0^2$.
Since $T_0 = mR\omega_0^2$,we have $T = 8T_0$.
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4
PhysicsMediumMCQAIIMS · 1985
$A$ wheel is subjected to uniform angular acceleration about its axis. Initially,its angular velocity is zero. In the first $2 \ s$,it rotates through an angle ${\theta _1}$. In the next $2 \ s$,it rotates through an additional angle ${\theta _2}$. The ratio of ${\theta _2}/{\theta _1}$ is:
A
$1$
B
$2$
C
$3$
D
$5$
Solution
(C) Given that the wheel starts from rest,so initial angular velocity ${\omega _0} = 0$. Let the uniform angular acceleration be $\alpha$.
Using the kinematic equation for angular displacement: $\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}$.
For the first $2 \ s$ $(t = 2 \ s)$: ${\theta _1} = 0 + \frac{1}{2}\alpha {(2)^2} = 2\alpha$ ... $(i)$.
For the total time of $4 \ s$ $(t = 2 + 2 = 4 \ s)$: ${\theta _1} + {\theta _2} = 0 + \frac{1}{2}\alpha {(4)^2} = 8\alpha$ ... $(ii)$.
Subtracting equation $(i)$ from equation $(ii)$: ${\theta _2} = 8\alpha - 2\alpha = 6\alpha$.
Now,the ratio is $\frac{{\theta _2}}{{\theta _1}} = \frac{6\alpha}{2\alpha} = 3$.
Using the kinematic equation for angular displacement: $\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}$.
For the first $2 \ s$ $(t = 2 \ s)$: ${\theta _1} = 0 + \frac{1}{2}\alpha {(2)^2} = 2\alpha$ ... $(i)$.
For the total time of $4 \ s$ $(t = 2 + 2 = 4 \ s)$: ${\theta _1} + {\theta _2} = 0 + \frac{1}{2}\alpha {(4)^2} = 8\alpha$ ... $(ii)$.
Subtracting equation $(i)$ from equation $(ii)$: ${\theta _2} = 8\alpha - 2\alpha = 6\alpha$.
Now,the ratio is $\frac{{\theta _2}}{{\theta _1}} = \frac{6\alpha}{2\alpha} = 3$.
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5
PhysicsMediumMCQAIIMS · 1985
Two planets have the same average density but their radii are $R_1$ and $R_2$. If acceleration due to gravity on these planets be $g_1$ and $g_2$ respectively,then
A
$\frac{g_1}{g_2} = \frac{R_1}{R_2}$
B
$\frac{g_1}{g_2} = \frac{R_2}{R_1}$
C
$\frac{g_1}{g_2} = \frac{R_1^2}{R_2^2}$
D
$\frac{g_1}{g_2} = \frac{R_1^3}{R_2^3}$
Solution
(A) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \rho \cdot V = \rho \cdot \frac{4}{3}\pi R^3$,we substitute this into the gravity formula:
$g = \frac{G}{R^2} \cdot (\rho \cdot \frac{4}{3}\pi R^3) = \frac{4}{3}\pi \rho GR$.
Given that the average density $\rho$ is the same for both planets,we have $g \propto R$.
Therefore,the ratio of the acceleration due to gravity on the two planets is $\frac{g_1}{g_2} = \frac{R_1}{R_2}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \rho \cdot V = \rho \cdot \frac{4}{3}\pi R^3$,we substitute this into the gravity formula:
$g = \frac{G}{R^2} \cdot (\rho \cdot \frac{4}{3}\pi R^3) = \frac{4}{3}\pi \rho GR$.
Given that the average density $\rho$ is the same for both planets,we have $g \propto R$.
Therefore,the ratio of the acceleration due to gravity on the two planets is $\frac{g_1}{g_2} = \frac{R_1}{R_2}$.
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6
PhysicsMediumMCQAIIMS · 1985
The value of Poisson's ratio lies between
A
$-1$ to $\frac{1}{2}$
B
$-\frac{3}{4}$ to $-\frac{1}{2}$
C
$-\frac{1}{2}$ to $1$
D
$1$ to $2$
Solution
(A) The relationship between Young's modulus $(Y)$,Bulk modulus $(K)$,and Modulus of rigidity $(\eta)$ with Poisson's ratio $(\sigma)$ is given by:
$Y = 3K(1 - 2\sigma)$
$Y = 2\eta(1 + \sigma)$
For a stable material,the moduli of elasticity must be positive $(Y, K, \eta > 0)$.
From $Y = 3K(1 - 2\sigma) > 0$,we get $1 - 2\sigma > 0$,which implies $\sigma < \frac{1}{2}$.
From $Y = 2\eta(1 + \sigma) > 0$,we get $1 + \sigma > 0$,which implies $\sigma > -1$.
Therefore,the theoretical range for Poisson's ratio is $-1 < \sigma < \frac{1}{2}$.
$Y = 3K(1 - 2\sigma)$
$Y = 2\eta(1 + \sigma)$
For a stable material,the moduli of elasticity must be positive $(Y, K, \eta > 0)$.
From $Y = 3K(1 - 2\sigma) > 0$,we get $1 - 2\sigma > 0$,which implies $\sigma < \frac{1}{2}$.
From $Y = 2\eta(1 + \sigma) > 0$,we get $1 + \sigma > 0$,which implies $\sigma > -1$.
Therefore,the theoretical range for Poisson's ratio is $-1 < \sigma < \frac{1}{2}$.
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7
PhysicsMediumMCQAIIMS · 1985
The molecules of a given mass of a gas have a $r.m.s.$ velocity of $200 \, m/s$ at $27^{\circ}C$ and $1.0 \times 10^5 \, N/m^2$ pressure. When the temperature is $127^{\circ}C$ and pressure is $0.5 \times 10^5 \, N/m^2$,the $r.m.s.$ velocity in $m/s$ will be
A
$\frac{100\sqrt{2}}{3}$
B
$100\sqrt{2}$
C
$\frac{400}{\sqrt{3}}$
D
None of the above
Solution
(C) The $r.m.s.$ velocity of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This shows that $v_{rms} \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.
Note that the $r.m.s.$ velocity is independent of the pressure of the gas.
Given:
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.
Initial $r.m.s.$ velocity $v_1 = 200 \, m/s$.
Final temperature $T_2 = 127^{\circ}C = 127 + 273 = 400 \, K$.
Using the proportionality $v_2 / v_1 = \sqrt{T_2 / T_1}$:
$v_2 = v_1 \times \sqrt{\frac{T_2}{T_1}}$
$v_2 = 200 \times \sqrt{\frac{400}{300}}$
$v_2 = 200 \times \sqrt{\frac{4}{3}}$
$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s$.
This shows that $v_{rms} \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.
Note that the $r.m.s.$ velocity is independent of the pressure of the gas.
Given:
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.
Initial $r.m.s.$ velocity $v_1 = 200 \, m/s$.
Final temperature $T_2 = 127^{\circ}C = 127 + 273 = 400 \, K$.
Using the proportionality $v_2 / v_1 = \sqrt{T_2 / T_1}$:
$v_2 = v_1 \times \sqrt{\frac{T_2}{T_1}}$
$v_2 = 200 \times \sqrt{\frac{400}{300}}$
$v_2 = 200 \times \sqrt{\frac{4}{3}}$
$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s$.
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8
PhysicsMediumMCQAIIMS · 1985
$E, m, l$ and $G$ denote energy,mass,angular momentum,and gravitational constant respectively. Then the dimensions of $\frac{El^2}{m^5G^2}$ are:
A
Angle
B
Length
C
Mass
D
Time
Solution
(A) The dimensional formulas for the given quantities are:
$[E] = [ML^2T^{-2}]$
$[m] = [M]$
$[l] = [ML^2T^{-1}]$
$[G] = [M^{-1}L^3T^{-2}]$
Substituting these dimensions into the expression $\frac{El^2}{m^5G^2}$:
$\frac{[ML^2T^{-2}] \cdot [ML^2T^{-1}]^2}{[M]^5 \cdot [M^{-1}L^3T^{-2}]^2} = \frac{[ML^2T^{-2}] \cdot [M^2L^4T^{-2}]}{[M^5] \cdot [M^{-2}L^6T^{-4}]} = \frac{[M^3L^6T^{-4}]}{[M^3L^6T^{-4}]} = [M^0L^0T^0]$
Since the dimensions are $[M^0L^0T^0]$,the quantity is dimensionless. Among the given options,an angle is a dimensionless quantity.
$[E] = [ML^2T^{-2}]$
$[m] = [M]$
$[l] = [ML^2T^{-1}]$
$[G] = [M^{-1}L^3T^{-2}]$
Substituting these dimensions into the expression $\frac{El^2}{m^5G^2}$:
$\frac{[ML^2T^{-2}] \cdot [ML^2T^{-1}]^2}{[M]^5 \cdot [M^{-1}L^3T^{-2}]^2} = \frac{[ML^2T^{-2}] \cdot [M^2L^4T^{-2}]}{[M^5] \cdot [M^{-2}L^6T^{-4}]} = \frac{[M^3L^6T^{-4}]}{[M^3L^6T^{-4}]} = [M^0L^0T^0]$
Since the dimensions are $[M^0L^0T^0]$,the quantity is dimensionless. Among the given options,an angle is a dimensionless quantity.
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