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CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 15 - MCQExams.com
CBSE
Class 11 Engineering Physics
Work,Energy And Power
Quiz 15
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
A body is falling a height $$h$$. After it has fallen a height $$\dfrac{h}{2}$$. it will possess
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only potential energy
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only kinetic energy
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half potential energy and half kinetic energy
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more kinetic and less potential energy
Explanation
$$(c)$$ half potential and half kinetic energy
Explanation: When the body is height $$h$$; its potential energy is at maximum and kinetic energy is zero. When the body hits the ground, its potential energy becomes zero and kinetic energy is at maximum. At mid-way i.e., half the height; its potential energy becomes half of the maximum potential energy and same happens to the kinetic energy.
Four alternatives are given to each of the following incomplete statements/questions. Choose the right answer:
Which of the following object has higher potential energy?
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mass $$ = 10\,kg$$ , $$ g = 9.8\, ms^{-2} , h = 10\,m $$
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mass $$ = 5\,kg , g = 9.8\, ms^{-2} , h = 12\,m $$
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mass $$ = 8\,kg , g = 9.8\,ms^{-2} , h = 100\,m $$
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mass $$ = 6\,kg , g = 9.8\,ms^{-2} , h = 20\,m $$
Explanation
The potential energy is given as-
$$PE = mgH$$
$$m$$; mass of the object
$$g = 9.8/ ms^{-2}$$
$$h$$; height gained from ground
For option A:
$$m = 10\ kg,\ h = 10\ m$$
$$PE = 10 \times 9.8 \times 10$$
$$\Rightarrow PE = 980\ J$$
For option B:
$$m = 5\ kg,\ h = 12\ m$$
$$PE = 5 \times 9.8 \times 12$$
$$\Rightarrow PE = 588\ J$$
For option C:
$$m = 8\ kg,\ h = 100\ m$$
$$PE = 8 \times 9.8 \times 100$$
$$\Rightarrow PE = 7840\ J$$
For option D:
$$m = 6\ kg,\ h = 20\ m$$
$$PE = 6 \times 9.8 \times 20$$
$$\Rightarrow PE = 1176\ J$$
For option C potential energy is highest so option C is the correct answer.
A force $$ F = K(yi + xj ) $$ ( where $$ K $$ is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is
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$$ -2 Ka^2 $$
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$$ 2K a^2 $$
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$$ -Ka^2 $$
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$$ Ka^2 $$
Explanation
While moving from$$ (0,0) to (a,0) $$
Along positive x-axis $$ y = 0 \therefore \overrightarrow {F} = -kx \hat {j} $$
i.e. force is in negative y-direction while displacement is in positive x-direction.
$$ \therefore W_1 = 0 $$
Because force is perpendicular to displacement
Then particle moves from $$ (a,0) $$ to $$ (a,a) $$ along a line parallel to y-axis $$ ( x = +a) $$ during this $$ \overrightarrow {F} = -k( y \hat {j} + a \hat {J}) $$
The first component of force $$ - ky \hat {i}$$ will not contribute any work because this component is along negative x-direction $$ ( - \hat {i}) $$ while displacement is in positive y- direction $$ (a,0) $$ to $$ (a,a) $$ The second component of force i.e. $$ - ka \hat {j} $$ will perform negative work.
$$ \therefore W_2 = ( - ka \hat {j})(a \hat {j}) = ( -ka)(a) = -ka^2 $$
So net work done on the particle $$ W = W_1 +W_2 $$
$$ = 0 + ( -ka^2) =-ka^2 $$
Is the work required to be done by an external force on an object on a frictionless, horizontal surface to accelerate it from a speed $$v$$ to a speed $$2v$$.
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equal to the work required to accelerate the object from $$v = 0$$ to $$v$$,
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twice the work required to accelerate the object from $$v = 0$$ to $$v$$
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three times the work required to accelerate the object from $$v = 0$$ to $$v$$,
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four times the work required to accelerate the object from $$0$$ to $$v$$
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not known without knowledge of the acceleration
Explanation
Answer (c). The net work needed to accelerate the object from $$v = 0$$ to $$v$$ is
$$W_{1}=KE_{1f}-KE_{1i}=\dfrac{1}{2}mv^{2}-\dfrac{1}{2}m(0)^{2}=\dfrac{1}{2}mv^{2}$$
The work required to accelerate the object from speed $$v$$ to speed $$2v$$ is
$$W_{2}=KE_{2f}-KE_{2i}=\dfrac{1}{2}m(2v)^{2}-\dfrac{1}{2}mv^{2}$$
$$=\dfrac{1}{2}m(4v^{2}-v^{2})=3\left ( \dfrac{1}{2}mv^{2} \right )=3W_{1}$$
Bullet $$2$$ has twice the mass of bullet $$1$$. Both are fired so that they have the same speed. If the kinetic energy of bullet $$1$$ is $$K$$, is the kinetic energy of bullet $$2$$
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$$0.25K$$
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$$0.5K$$
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$$0.71K$$
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$$K$$
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$$2K$$
Explanation
Let $$m$$ be the mass of bullet $$1$$.
According to the given condition, the mass of bullet $$2$$ is $$2m.$$
Kinetic energy of bullet $$1$$ is
$$KE_1=\dfrac{1}{2}mv^2=K$$
KE of bullet $$2$$ is
$$KE_2=\dfrac{1}{2}(2m)v^2=2(\dfrac{1}{2}mv^2)=2K$$
The displacement of a particle of mass $$m=2\,kg$$ on a smooth horizontal surface is a function of time given by $$x=e^{2t}$$. Find net the work done by external agent from $$t=\,ln\,1$$ to $$t=\,ln\,2$$.
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$$20\,J$$
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$$30\,J$$
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$$50\,J$$
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$$60\,J$$
Explanation
Displacement of particle $$x=e^{2t}$$
$$\therefore$$ velocity at any instant $$'t'$$ is given by $$v=\dfrac{dx}{dt}=2e^{2t}$$
$$\therefore$$ acceleration at any instant $$'t'$$ is given by
$$a=\dfrac{dv}{dt}=4e^{2t}=\dfrac{d^2x}{dt^2}$$
work done by the force $$\displaystyle W=\int F.dx=\int F.\dfrac{dx}{dt}.dt$$
$$\displaystyle W= \int^{ln\,2}_{ln\,1}ma\left(\dfrac{dx}{dt}\right)dt$$ $$m=2kg$$
$$\displaystyle =\int^{ln\,2}_{ln\,1}2\times 4e^{2t}\times 2e^{2t}dt$$
$$\displaystyle =16\int^{ln\,2}_{ln\,1}e^{4t}dt=\dfrac{16}{4}[e^{4t}]^{ln\,2}_{ln\,1}$$
$$=\dfrac{16}{4}[e^{4\,ln\,2}-e^{4\,ln\,1}]$$
$$=4[e^{ln\,(2^4)}-e^{ln\,(1^4)}]$$
$$=4[2^4-1^4]=4[15]=60\,Joules$$
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