Work Energy Theorem, Power and Vertical Circle — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Work Energy Theorem, Power and Vertical Circle MCQs with step-by-step solutions (16 questions). Part of Work, Energy and Power. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Work Energy Theorem, Power and Vertical Circle · medium
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? $(g=10$ m/s$^2)$
A. 10.2 kW
B. 8.1 kW ✓ Correct
C. 12.3 kW
D. 7.0 kW
Solution: Flow rate $\dfrac{m}{t}=15$ kg/s, $h=60$ m, $g=10$ m/s$^2$. With a 10% loss, 90% of the input is used, so power generated $=0.90\times\dfrac{mgh}{t}=0.90\times15\times10\times60=8100$ W $=8.1$ kW.
Q2 — Work Energy Theorem, Power and Vertical Circle · medium
A point mass $m$ is moved in a vertical circle of radius $r$ with the help of a string. The velocity of the mass is $\sqrt{7gr}$ at the lowest point. The tension in the string at the lowest point is
A. $6mg$
B. $7mg$
C. $8mg$ ✓ Correct
D. $1mg$
Solution: At the lowest point $v=\sqrt{7gr}>\sqrt{5gr}$, so the mass completes the vertical circle. Here $T_{bottom}-mg=\dfrac{mv^2}{r}=\dfrac{m}{r}(\sqrt{7gr})^2=7mg\Rightarrow T_{bottom}=8mg$.
Q3 — Work Energy Theorem, Power and Vertical Circle · medium
An object of mass 500 g, initially at rest acted upon by a variable force whose $X$ component varies with $X$ in the manner shown. The velocities of the object at points $X=8$ m and $X=12$ m, would be the respective values of (nearly)
A. 18 m/s and 24.4 m/s
B. 23 m/s and 24.4 m/s
C. 23 m/s and 20.6 m/s ✓ Correct
D. 18 m/s and 20.6 m/s
Solution: The area under the force-displacement curve gives the work done. From the work-energy theorem, $W=\Delta KE$. At $x=8$ m, $W=\text{Area } ABDO+\text{Area } CEFD=20\times5+10\times3=130$ J, so $130=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times\dfrac{500}{1000}v^2\Rightarrow v=2\sqrt{130}=22.8\approx23$ ms$^{-1}$. At $x=12$ m, $W=20\times5+10\times3+(-20\times2)+\left(\dfrac{1}{2}\times-5\times2\right)+10\times2=105$ J, so $105=\dfrac{1}{2}\times\dfrac{1}{2}\times v^2\Rightarrow v=2\sqrt{105}\approx20.6$ ms$^{-1}$.
Q4 — Work Energy Theorem, Power and Vertical Circle · medium
A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
A. the wire is horizontal
B. the mass is at the lowest point ✓ Correct
C. inclined at an angle of $60^\circ$ from vertical
D. the mass is at the highest point
Solution: At any point $P$ the net force towards the centre is centripetal, so $T-mg\cos\theta=\dfrac{mv^2}{l}\Rightarrow T=mg\cos\theta+\dfrac{mv^2}{l}$. At the lowest point ($\theta=0^\circ$), $T_A=mg+\dfrac{mv^2}{l}$, which is the maximum; at the top ($\theta=180^\circ$), $T_C=-mg+\dfrac{mv^2}{l}$. Tension is maximum at the lowest point, so the wire is most likely to break there.
Q5 — Work Energy Theorem, Power and Vertical Circle · medium
A body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) just completes a vertical circle of diameter $AB=D$. The height $h$ is equal to
A. $\dfrac{7}{5}D$
B. $D$
C. $\dfrac{3}{2}D$
D. $\dfrac{5}{4}D$ ✓ Correct
Solution: On a frictionless surface mechanical energy is conserved: $(KE)_i+(PE)_i=(KE)_f+(PE)_f\Rightarrow 0+mgh=\dfrac{1}{2}mv_A^2+0$, so $h=\dfrac{v_A^2}{2g}$. To complete the vertical circle, $v_A=v_{min}=\sqrt{5gR}$ with $R=\dfrac{AB}{2}=\dfrac{D}{2}$, giving $v_A=\sqrt{\dfrac{5}{2}gD}$. Substituting, $h=\dfrac{\left(\sqrt{\dfrac{5}{2}gD}\right)^2}{2g}=\dfrac{5gD}{2\times2g}=\dfrac{5}{4}D$.
Q6 — Work Energy Theorem, Power and Vertical Circle · medium
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take $g$ constant with a value of 10 m/s$^2$. The work done by the (i) gravitational force and the (ii) resistive force of air is
A. (i) $-10$ J, (ii) $-8.25$ J
B. (i) $1.25$ J, (ii) $-8.25$ J
C. (i) $100$ J, (ii) $8.75$ J
D. (i) $10$ J, (ii) $-8.75$ J ✓ Correct
Solution: By the work-KE theorem, the change in KE equals the work done by all forces. Work done by gravity, $W_g=mgh=10^{-3}\times10\times1\times10^3=10$ J. Then $\Delta K=W_{gravity}+W_{air}\Rightarrow \dfrac{1}{2}mv^2=mgh+W_{air}\Rightarrow W_{air}=\dfrac{1}{2}mv^2-mgh=10^{-3}\left(\dfrac{1}{2}\times50\times50-10\times10^3\right)=-8.75$ J.
Q7 — Work Energy Theorem, Power and Vertical Circle · medium
A body of mass 1 kg begins to move under the action of a time dependent force $\mathbf{F}=(2t\hat{\mathbf{i}}+3t^2\hat{\mathbf{j}})$ N, where $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along $X$ and $Y$ axes. What power will be developed by the force at the time $(t)$?
A. $(2t^2+4t^4)$ W
B. $(2t^3+3t^4)$ W
C. $(2t^3+3t^5)$ W ✓ Correct
D. $(2t+3t^3)$ W
Solution: $\mathbf{F}=m\mathbf{a}\Rightarrow \mathbf{a}=\dfrac{\mathbf{F}}{m}=(2t\hat{\mathbf{i}}+3t^2\hat{\mathbf{j}})$ m/s$^2$ (since $m=1$ kg). As $\mathbf{a}=\dfrac{d\mathbf{v}}{dt}$, integrating gives $\mathbf{v}=t^2\hat{\mathbf{i}}+t^3\hat{\mathbf{j}}$. Power $P=\mathbf{F}\cdot\mathbf{v}=(2t\hat{\mathbf{i}}+3t^2\hat{\mathbf{j}})\cdot(t^2\hat{\mathbf{i}}+t^3\hat{\mathbf{j}})=2t^3+3t^5$ W.
Q8 — Work Energy Theorem, Power and Vertical Circle · medium
What is the minimum velocity with which a body of mass $m$ must enter a vertical loop of radius $R$ so that it can complete the loop?
A. $\sqrt{2gR}$
B. $\sqrt{3gR}$
C. $\sqrt{5gR}$ ✓ Correct
D. $\sqrt{gR}$
Solution: At the top point $C$, Newton's second law gives $T_c+mg=\dfrac{mv_c^2}{R}$. To just complete the loop, $T_c\geq0$, so $mg=\dfrac{mv_c^2}{R}\Rightarrow v_c=\sqrt{gR}$. By conservation of energy between the bottom $A$ and top $C$: $\dfrac{1}{2}mv_0^2=\dfrac{1}{2}mv_c^2+2mgR\Rightarrow v_0^2=gR+4gR=5gR\Rightarrow v_0=\sqrt{5gR}$.
Q9 — Work Energy Theorem, Power and Vertical Circle · medium
A particle of mass $m$ is driven by a machine that delivers a constant power $k$ watts. If the particle starts from rest, the force on the particle at time $t$ is
A. $\sqrt{\dfrac{mk}{2}}\,t^{-1/2}$ ✓ Correct
B. $\sqrt{mk}\,t^{-1/2}$
C. $\sqrt{2mk}\,t^{-1/2}$
D. $\dfrac{1}{2}\sqrt{mk}\,t^{-1/2}$
Solution: Constant power means $F\cdot v=k$, so $m\dfrac{dv}{dt}v=k\Rightarrow \int v\,dv=\dfrac{k}{m}\int dt\Rightarrow \dfrac{v^2}{2}=\dfrac{k}{m}t\Rightarrow v=\sqrt{\dfrac{2kt}{m}}$. Then $F=m\dfrac{dv}{dt}=m\dfrac{d}{dt}\left(\dfrac{2kt}{m}\right)^{1/2}=\sqrt{2km}\cdot\dfrac{1}{2}t^{-1/2}=\sqrt{\dfrac{mk}{2}}\,t^{-1/2}$.
Q10 — Work Energy Theorem, Power and Vertical Circle · medium
The heart of a man pumps 5 L of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be $13.6\times10^3$ kg/m$^3$ and $g=10$ m/s$^2$, then the power of heart in watt is
A. $1.70$ ✓ Correct
B. $2.35$
C. $3.0$
D. $1.50$
Solution: Pumping rate $\dfrac{dV}{dt}=\dfrac{5\times10^{-3}}{60}$ m$^3$/s. Power $=P\dfrac{dV}{dt}=\rho gh\dfrac{dV}{dt}=\dfrac{(13.6\times10^3)(10\times0.15\times5\times10^{-3})}{60}=1.70$ W.
Q11 — Work Energy Theorem, Power and Vertical Circle · medium
An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of $2$ m s$^{-1}$. The mass per unit length of water in the pipe is 100 kg m$^{-1}$. What is the power of the engine?
A. 400 W
B. 200 W
C. 100 W
D. 800 W ✓ Correct
Solution: Velocity of water $v=2$ m/s and mass per unit length $\dfrac{m}{l}=100$ kg/m. Power $=\dfrac{m}{l}\times v^3=100\times2\times2\times2=800$ W.
Q12 — Work Energy Theorem, Power and Vertical Circle · medium
An engine pumps water continuously through a hose. Water leaves the hose with a velocity $v$ and $m$ is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
A. $\dfrac{1}{2}mv^3$ ✓ Correct
B. $mv^3$
C. $\dfrac{1}{2}mv^2$
D. $\dfrac{1}{2}m^2v^2$
Solution: As $m$ is the mass per unit length, the rate of mass per second $=\dfrac{mx}{t}=mv$. $\therefore$ Rate of KE $=\dfrac{1}{2}(mv)v^2=\dfrac{1}{2}mv^3$.
Q13 — Work Energy Theorem, Power and Vertical Circle · medium
Water falls from a height of $60$ m at the rate of $15$ kg/s to operate a turbine. The losses due to frictional forces are $10\%$ of energy. How much power is generated by the turbine? (Take $g=10$ m/s$^2$)
A. $8.1$ kW ✓ Correct
B. $10.2$ kW
C. $12.3$ kW
D. $7.0$ kW
Solution: $P_{generated}=P_{input}\times\dfrac{90}{100}=\dfrac{mgh}{t}\times\dfrac{90}{100}=\dfrac{15\times10\times60}{1}\times\dfrac{90}{100}=8.1$ kW.
Q14 — Work Energy Theorem, Power and Vertical Circle · medium
A stone is tied to a string of length $l$ and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in velocity as it reaches a position where the string is horizontal ($g$ being acceleration due to gravity) is
A. $\sqrt{2(u^2-gl)}$ ✓ Correct
B. $\sqrt{u^2-gl}$
C. $u-\sqrt{u^2-2gl}$
D. $\sqrt{2gl}$
Solution: At the lowest position the stone has only KE, $K=\dfrac{1}{2}mu^2$. At the horizontal position, $E=U+K=\dfrac{1}{2}mu'^2+mgl$. By conservation of mechanical energy, $\dfrac{1}{2}mu^2=\dfrac{1}{2}mu'^2+mgl$, so $u'^2=u^2-2gl$, i.e. $u'=\sqrt{u^2-2gl}$. The two velocities are mutually perpendicular, so $|\Delta u|=\sqrt{u'^2+u^2+2u'u\cos90^\circ}=\sqrt{u'^2+u^2}=\sqrt{2(u^2-gl)}$.
Q15 — Work Energy Theorem, Power and Vertical Circle · medium
A stone is attached to one end of a string and rotated in a vertical circle. If string breaks at the position of maximum tension, it will break at
A. A
B. B ✓ Correct
C. C
D. D
Solution: When the string makes an angle $\theta$ with the vertical, balancing forces gives $T-mg\cos\theta=\dfrac{mv^2}{l}$, or $T=mg\cos\theta+\dfrac{mv^2}{l}$. Tension is maximum when $\cos\theta=+1$, i.e. $\theta=0$. This occurs at the lowest point $B$, so the string breaks at $B$.
Q16 — Work Energy Theorem, Power and Vertical Circle · medium
How much water a pump of $2$ kW can raise in one minute to a height of $10$ m? (Take $g=10$ m/s$^2$)
A. $1000$ L
B. $1200$ L ✓ Correct
C. $100$ L
D. $2000$ L
Solution: Power $=\dfrac{work}{time}=\dfrac{W}{t}$. Given $P=2$ kW $=2000$ W, $W=Mgh=M\times10\times10=100M$ and $t=60$ s. $\therefore 2000=\dfrac{100M}{60}$, so $M=1200$ kg and $V=1200$ L.