PhysicsQ201–210 of 399 questions
Page 5 of 5 · English
201
PhysicsMediumMCQAP EAMCET · 2025
$A$ body is moving along a straight line under the influence of a constant power source. If the relation between the displacement $(s)$ of the body and time $(t)$ is $s \propto t^x$, then $x=$
A
$1$
B
$2$
C
$\frac{2}{3}$
D
$\frac{3}{2}$
Solution
(D) Power $(P)$ is defined as the rate of doing work, $P = Fv$. Since $P$ is constant, $Fv = \text{constant}$.
Using Newton's second law, $F = ma = m \frac{dv}{dt}$.
Thus, $m \frac{dv}{dt} v = P$, which implies $v dv = \frac{P}{m} dt$.
Integrating both sides, $\int v dv = \int \frac{P}{m} dt$, we get $\frac{1}{2} v^2 = \frac{P}{m} t + C$.
Assuming the body starts from rest at $t=0$, $C=0$, so $v^2 = \frac{2P}{m} t$, or $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$, we have $\frac{ds}{dt} = k t^{1/2}$ where $k = \sqrt{\frac{2P}{m}}$.
Integrating with respect to time, $s = \int k t^{1/2} dt = k \frac{t^{3/2}}{3/2} = \frac{2k}{3} t^{3/2}$.
Therefore, $s \propto t^{3/2}$, which means $x = \frac{3}{2}$.
Using Newton's second law, $F = ma = m \frac{dv}{dt}$.
Thus, $m \frac{dv}{dt} v = P$, which implies $v dv = \frac{P}{m} dt$.
Integrating both sides, $\int v dv = \int \frac{P}{m} dt$, we get $\frac{1}{2} v^2 = \frac{P}{m} t + C$.
Assuming the body starts from rest at $t=0$, $C=0$, so $v^2 = \frac{2P}{m} t$, or $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$, we have $\frac{ds}{dt} = k t^{1/2}$ where $k = \sqrt{\frac{2P}{m}}$.
Integrating with respect to time, $s = \int k t^{1/2} dt = k \frac{t^{3/2}}{3/2} = \frac{2k}{3} t^{3/2}$.
Therefore, $s \propto t^{3/2}$, which means $x = \frac{3}{2}$.
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202
PhysicsMediumMCQAP EAMCET · 2025
$A$ crane of efficiency $80 \%$ is used to lift $8000 \ kg$ of coal from a mine of depth $108 \ m$. If the time taken by the crane to lift the coal is one hour, then the power of the crane (in $kW$) is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)
A
$5$
B
$4$
C
$6$
D
$3$
Solution
(D) The work done $(W)$ to lift the coal is given by $W = mgh$, where $m = 8000 \ kg$, $g = 10 \ m \ s^{-2}$, and $h = 108 \ m$.
$W = 8000 \times 10 \times 108 = 8.64 \times 10^6 \ J$.
The time taken $(t)$ is $1 \ hour = 3600 \ s$.
The useful power output $(P_{out})$ is $P_{out} = W / t = (8.64 \times 10^6) / 3600 = 2400 \ W = 2.4 \ kW$.
Given the efficiency $(\eta)$ is $80 \% = 0.8$, the input power $(P_{in})$ is $P_{in} = P_{out} / \eta$.
$P_{in} = 2.4 / 0.8 = 3 \ kW$.
Thus, the power of the crane is $3 \ kW$.
$W = 8000 \times 10 \times 108 = 8.64 \times 10^6 \ J$.
The time taken $(t)$ is $1 \ hour = 3600 \ s$.
The useful power output $(P_{out})$ is $P_{out} = W / t = (8.64 \times 10^6) / 3600 = 2400 \ W = 2.4 \ kW$.
Given the efficiency $(\eta)$ is $80 \% = 0.8$, the input power $(P_{in})$ is $P_{in} = P_{out} / \eta$.
$P_{in} = 2.4 / 0.8 = 3 \ kW$.
Thus, the power of the crane is $3 \ kW$.
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203
PhysicsEasyMCQAP EAMCET · 2025
$A$ body of mass '$M$' is moving with a uniform speed of '$v$' on a frictionless horizontal surface under the influence of two forces $F_1$ and $F_2$ as shown in the figure. The net power of the system is
A
$(F_1-F_2) v$
B
$0.5(F_1+F_2) v$
C
$(F_1+F_2) v$
D
Zero
Solution
(D) Power is defined as the dot product of force and velocity,given by $P = \vec{F} \cdot \vec{v}$.
Since the body is moving with a uniform speed '$v$',the net force acting on the body must be zero according to Newton's first law of motion.
From the figure,the forces $F_1$ and $F_2$ act in opposite directions. Therefore,the net force $F_{net} = F_1 - F_2 = 0$.
The net power $P_{net}$ is the power delivered by the net force:
$P_{net} = F_{net} \cdot v = (F_1 - F_2) \cdot v$.
Since $F_1 = F_2$ for uniform motion,$P_{net} = 0 \cdot v = 0$.
Since the body is moving with a uniform speed '$v$',the net force acting on the body must be zero according to Newton's first law of motion.
From the figure,the forces $F_1$ and $F_2$ act in opposite directions. Therefore,the net force $F_{net} = F_1 - F_2 = 0$.
The net power $P_{net}$ is the power delivered by the net force:
$P_{net} = F_{net} \cdot v = (F_1 - F_2) \cdot v$.
Since $F_1 = F_2$ for uniform motion,$P_{net} = 0 \cdot v = 0$.
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204
PhysicsMediumMCQAP EAMCET · 2025
$A$ motor can pump $7560 \text{ kg}$ of water per hour from a well of depth $100 \text{ m}$. If the efficiency of the pump is $70 \%$, then the power of the pump is (Acceleration due to gravity $= 10 \text{ m s}^{-2}$) (in $\text{ kW}$)
A
$4$
B
$6$
C
$3$
D
$7$
Solution
(C) The mass of water pumped per hour is $m = 7560 \text{ kg}$.
The time taken is $t = 1 \text{ hour} = 3600 \text{ s}$.
The depth of the well is $h = 100 \text{ m}$.
The acceleration due to gravity is $g = 10 \text{ m s}^{-2}$.
The useful work done (output energy) per second is the power output $P_{\text{out}} = \frac{mgh}{t}$.
$P_{\text{out}} = \frac{7560 \times 10 \times 100}{3600} = \frac{7560000}{3600} = 2100 \text{ W} = 2.1 \text{ kW}$.
Given the efficiency $\eta = 70 \% = 0.7$, the power input $P_{\text{in}}$ is given by $\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$.
$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2.1 \text{ kW}}{0.7} = 3 \text{ kW}$.
Therefore, the power of the pump is $3 \text{ kW}$.
The time taken is $t = 1 \text{ hour} = 3600 \text{ s}$.
The depth of the well is $h = 100 \text{ m}$.
The acceleration due to gravity is $g = 10 \text{ m s}^{-2}$.
The useful work done (output energy) per second is the power output $P_{\text{out}} = \frac{mgh}{t}$.
$P_{\text{out}} = \frac{7560 \times 10 \times 100}{3600} = \frac{7560000}{3600} = 2100 \text{ W} = 2.1 \text{ kW}$.
Given the efficiency $\eta = 70 \% = 0.7$, the power input $P_{\text{in}}$ is given by $\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$.
$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2.1 \text{ kW}}{0.7} = 3 \text{ kW}$.
Therefore, the power of the pump is $3 \text{ kW}$.
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205
PhysicsMediumMCQAP EAMCET · 2025
If a force of $(6x^2 - 4x) \ N$ acts on a body of mass $10 \ kg$,then the work done by the force in displacing the body from $x = 2 \ m$ to $x = 4 \ m$ is: (in $J$)
A
$22$
B
$44$
C
$66$
D
$88$
Solution
(D) The work done $W$ by a variable force $F(x)$ is given by the integral $W = \int_{x_1}^{x_2} F(x) \ dx$.
Given $F(x) = 6x^2 - 4x$,$x_1 = 2 \ m$,and $x_2 = 4 \ m$.
$W = \int_{2}^{4} (6x^2 - 4x) \ dx$.
Integrating the expression: $W = [\frac{6x^3}{3} - \frac{4x^2}{2}]_{2}^{4} = [2x^3 - 2x^2]_{2}^{4}$.
Substituting the limits: $W = (2(4)^3 - 2(4)^2) - (2(2)^3 - 2(2)^2)$.
$W = (2(64) - 2(16)) - (2(8) - 2(4))$.
$W = (128 - 32) - (16 - 8)$.
$W = 96 - 8 = 88 \ J$.
Given $F(x) = 6x^2 - 4x$,$x_1 = 2 \ m$,and $x_2 = 4 \ m$.
$W = \int_{2}^{4} (6x^2 - 4x) \ dx$.
Integrating the expression: $W = [\frac{6x^3}{3} - \frac{4x^2}{2}]_{2}^{4} = [2x^3 - 2x^2]_{2}^{4}$.
Substituting the limits: $W = (2(4)^3 - 2(4)^2) - (2(2)^3 - 2(2)^2)$.
$W = (2(64) - 2(16)) - (2(8) - 2(4))$.
$W = (128 - 32) - (16 - 8)$.
$W = 96 - 8 = 88 \ J$.
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206
PhysicsEasyMCQAP EAMCET · 2025
If a force $\vec{F}=(3 \hat{i}-2 \hat{j}) \text{ N}$ acting on a body displaces it from point $(1 \text{ m}, 2 \text{ m})$ to point $(2 \text{ m}, 0 \text{ m})$,then the work done by the force is (in $\text{ J}$)
A
$5$
B
$6$
C
$4$
D
$7$
Solution
(D) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of the force vector and the displacement vector $\vec{d}$.
Given $\vec{F} = (3 \hat{i} - 2 \hat{j}) \text{ N}$.
The initial position vector is $\vec{r}_1 = (1 \hat{i} + 2 \hat{j}) \text{ m}$.
The final position vector is $\vec{r}_2 = (2 \hat{i} + 0 \hat{j}) \text{ m}$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (2-1) \hat{i} + (0-2) \hat{j} = (1 \hat{i} - 2 \hat{j}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (3 \hat{i} - 2 \hat{j}) \cdot (1 \hat{i} - 2 \hat{j})$.
$W = (3 \times 1) + (-2 \times -2) = 3 + 4 = 7 \text{ J}$.
Given $\vec{F} = (3 \hat{i} - 2 \hat{j}) \text{ N}$.
The initial position vector is $\vec{r}_1 = (1 \hat{i} + 2 \hat{j}) \text{ m}$.
The final position vector is $\vec{r}_2 = (2 \hat{i} + 0 \hat{j}) \text{ m}$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (2-1) \hat{i} + (0-2) \hat{j} = (1 \hat{i} - 2 \hat{j}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (3 \hat{i} - 2 \hat{j}) \cdot (1 \hat{i} - 2 \hat{j})$.
$W = (3 \times 1) + (-2 \times -2) = 3 + 4 = 7 \text{ J}$.
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207
PhysicsMediumMCQAP EAMCET · 2025
If a position-dependent force $F(x) = (3x^2 - 2x + 7) \text{ N}$ acting on a body of mass $2 \text{ kg}$ displaces it from $x = 0 \text{ m}$ to $x = 5 \text{ m}$,then the work done by the force is: (in $J$)
A
$165$
B
$115$
C
$150$
D
$135$
Solution
(D) The work done $W$ by a variable force $F(x)$ is given by the integral of the force with respect to displacement:
$W = \int_{x_1}^{x_2} F(x) \, dx$
Given $F(x) = 3x^2 - 2x + 7$,$x_1 = 0 \text{ m}$,and $x_2 = 5 \text{ m}$.
$W = \int_{0}^{5} (3x^2 - 2x + 7) \, dx$
Integrating term by term:
$W = [x^3 - x^2 + 7x]_{0}^{5}$
Substitute the limits:
$W = (5^3 - 5^2 + 7(5)) - (0^3 - 0^2 + 7(0))$
$W = (125 - 25 + 35) - 0$
$W = 100 + 35 = 135 \text{ J}$.
Thus,the work done is $135 \text{ J}$.
$W = \int_{x_1}^{x_2} F(x) \, dx$
Given $F(x) = 3x^2 - 2x + 7$,$x_1 = 0 \text{ m}$,and $x_2 = 5 \text{ m}$.
$W = \int_{0}^{5} (3x^2 - 2x + 7) \, dx$
Integrating term by term:
$W = [x^3 - x^2 + 7x]_{0}^{5}$
Substitute the limits:
$W = (5^3 - 5^2 + 7(5)) - (0^3 - 0^2 + 7(0))$
$W = (125 - 25 + 35) - 0$
$W = 100 + 35 = 135 \text{ J}$.
Thus,the work done is $135 \text{ J}$.
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208
PhysicsEasyMCQAP EAMCET · 2025
If a constant force of $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$ acting on a body of mass $5 \text{ kg}$ displaces it from $(3 \hat{i} - 4 \hat{k}) \text{ m}$ to $(2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$, then the work done by the force on the body is (in $\text{ J}$)
A
$32$
B
$28$
C
$36$
D
$44$
Solution
(A) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of the force vector and the displacement vector $\vec{d}$.
Given force $\vec{F} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$.
The initial position vector is $\vec{r}_1 = (3 \hat{i} - 4 \hat{k}) \text{ m}$.
The final position vector is $\vec{r}_2 = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (2 - 3) \hat{i} + (2 - 0) \hat{j} + (3 - (-4)) \hat{k} = (-1 \hat{i} + 2 \hat{j} + 7 \hat{k}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (-1 \hat{i} + 2 \hat{j} + 7 \hat{k})$.
$W = (2 \times -1) + (3 \times 2) + (4 \times 7) = -2 + 6 + 28 = 32 \text{ J}$.
Given force $\vec{F} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$.
The initial position vector is $\vec{r}_1 = (3 \hat{i} - 4 \hat{k}) \text{ m}$.
The final position vector is $\vec{r}_2 = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (2 - 3) \hat{i} + (2 - 0) \hat{j} + (3 - (-4)) \hat{k} = (-1 \hat{i} + 2 \hat{j} + 7 \hat{k}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (-1 \hat{i} + 2 \hat{j} + 7 \hat{k})$.
$W = (2 \times -1) + (3 \times 2) + (4 \times 7) = -2 + 6 + 28 = 32 \text{ J}$.
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209
PhysicsEasyMCQAP EAMCET · 2025
If a force $(3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \text{ N}$ acting on a body displaces it through $(2 \hat{i} + 2 \hat{j} + 1 \hat{k}) \text{ m}$,then the work done by the force on the body is (in $J$)
A
$40$
B
$20$
C
$15$
D
$25$
Solution
(C) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{d}$ is given by the dot product of the force and displacement vectors:
$W = \vec{F} \cdot \vec{d}$
Given:
$\vec{F} = (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \text{ N}$
$\vec{d} = (2 \hat{i} + 2 \hat{j} + 1 \hat{k}) \text{ m}$
Calculating the dot product:
$W = (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} + 1 \hat{k})$
$W = (3 \times 2) + (2 \times 2) + (5 \times 1)$
$W = 6 + 4 + 5$
$W = 15 \text{ J}$
Therefore,the work done is $15 \text{ J}$.
$W = \vec{F} \cdot \vec{d}$
Given:
$\vec{F} = (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \text{ N}$
$\vec{d} = (2 \hat{i} + 2 \hat{j} + 1 \hat{k}) \text{ m}$
Calculating the dot product:
$W = (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} + 1 \hat{k})$
$W = (3 \times 2) + (2 \times 2) + (5 \times 1)$
$W = 6 + 4 + 5$
$W = 15 \text{ J}$
Therefore,the work done is $15 \text{ J}$.
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210
PhysicsMediumMCQAP EAMCET · 2025
The power required for an engine to maintain a constant speed of $50 \,m \,s^{-1}$ for a train of mass $3 \times 10^6 \,kg$ on rough rails is (The coefficient of kinetic friction between the rails and wheels of the train is $0.05$ and acceleration due to gravity $= 10 \,m \,s^{-2}$).
A
$75 \,MW$
B
$40 \,MW$
C
$75 \,kW$
D
$65 \,MW$
Solution
(A) To maintain a constant speed, the engine must exert a force equal to the frictional force acting on the train.
The frictional force $F_f$ is given by $F_f = \mu N$, where $\mu$ is the coefficient of kinetic friction and $N$ is the normal force.
Since the train is on a horizontal track, $N = mg$.
Given: $m = 3 \times 10^6 \,kg$, $\mu = 0.05$, $g = 10 \,m \,s^{-2}$, and velocity $v = 50 \,m \,s^{-1}$.
$F_f = 0.05 \times (3 \times 10^6 \,kg) \times (10 \,m \,s^{-2}) = 0.05 \times 3 \times 10^7 \,N = 1.5 \times 10^6 \,N$.
The power $P$ required is given by $P = F_f \times v$.
$P = (1.5 \times 10^6 \,N) \times (50 \,m \,s^{-1}) = 75 \times 10^6 \,W = 75 \,MW$.
The frictional force $F_f$ is given by $F_f = \mu N$, where $\mu$ is the coefficient of kinetic friction and $N$ is the normal force.
Since the train is on a horizontal track, $N = mg$.
Given: $m = 3 \times 10^6 \,kg$, $\mu = 0.05$, $g = 10 \,m \,s^{-2}$, and velocity $v = 50 \,m \,s^{-1}$.
$F_f = 0.05 \times (3 \times 10^6 \,kg) \times (10 \,m \,s^{-2}) = 0.05 \times 3 \times 10^7 \,N = 1.5 \times 10^6 \,N$.
The power $P$ required is given by $P = F_f \times v$.
$P = (1.5 \times 10^6 \,N) \times (50 \,m \,s^{-1}) = 75 \times 10^6 \,W = 75 \,MW$.
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