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CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 15 - MCQExams.com
CBSE
Class 11 Medical Physics
Work, Energy And Power
Quiz 15
Three identical blocks $$A,B$$ and $$C$$ are placed on horizontal frictionless surface. The blocks $$B$$ and $$C$$ are at rest. But $$A$$ is approaching towards $$B$$ with a speed $$10\ m/s$$. The coefficient of restitution for all collision is $$0.5$$. The speed of the block $$C$$ just after collision is:
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$$5.6\ m/s$$
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$$6\ m/s$$
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$$8\ m/s$$
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$$10\ m/s$$
The graph below show the force acting on a particle moves along the positive $$x-$$axis from the origin to $$x=x_1$$. The force if parallel to the $$x-$$axis and is conservative. The maximum magnitude $$F_1$$ has the same values for all graphs. Rank the solutions according to the change in the potential energy associated with the force. least (or most negative) to greatest (or most positive)
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$$2, 1, 3$$
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$$1, 3, 2$$
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$$2, 3, 1$$
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$$3, 2, 1$$
A particle of mass m moves in conservative force field along x axis where the potential energy U varies with position coordinate x as $$ U = U_0 ( -cos ax) , U_0 $$ and a being positive constants .which of the following statements is true regarding its motion. its total energy is $$ U_0 $$ and starts from x =
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the acceleration is constant
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it's speed is maximum at the initial position
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it's maximum x coordinate is $$ \frac { \pi}{ 2a} $$
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It's maximum kinetic energy is $$ U_0 $$
After how much time collision between the blocks will not take place practically:
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$$8\ sec$$
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$$16\ sec$$
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$$12\ sec$$
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$$infinite$$
A cannon of mass $$1000$$ kg located at the base of an inclined plane fires a shell of mass $$50$$ kg in horizontal direction with velocity $$180$$ km/h. The angle of inclination of the inclined plane with the horizontal is $$45^o$$. The coefficient of friction between the cannon and inclined plane is $$0.5$$. The maximum height, in metre, to which the cannon can ascend the inclined plane as a result of recoil is?
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$$\dfrac{5}{6}$$
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$$\dfrac{5}{24}$$
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$$\dfrac{5}{12}$$
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None of these
Explanation
Velocity of shell$$=180$$ km/h$$=50$$m/s
Let V be the velocity of the cannon after firing, then from the conservation of linear momentum,
$$0=50\times 50-1000V\Rightarrow V=2.5m/s$$
Let the cannon go up by height h.
Net work done by external forces $$=\Delta KE$$
$$\Rightarrow -Mgh-\mu Mg\cos\theta \times \dfrac{h}{\sin\theta}=0-\dfrac{1}{2}MV$$
where M is the mass of cannon,
Solving, we get $$h=5/24$$ m.
A wad of sticky clay of mass $$m$$ and velocity $$v_{i}$$ is fired at a solid cylinder of mass $$M$$ and radius $$R$$ figure. The cylinder if initially at rest and is mounted on a fixed horizontal axle that runs through the centre of mass. The line of motion of the projectile is perpendicular to the axle and at a distance $$d$$, less than $$R$$, from the centre.
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Angular velocity just after collision is $$\omega=\dfrac{2mv_{1}d}{(M+2m)R^{2}}$$
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Linear momentum of cylinder and clay is conserved
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Angular momentum of cylinder and clay is conserved
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Mechanical energy is conserved
A ball moving with a velocity v hits a massive wall moving towards the ball with a velocity u. An elastic impact lasts for time $$\Delta t$$.
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The average elastic force acting on the ball is $$[m(u+v)]/\Delta t$$
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The average elastic force on the ball is $$[2m(u+v)]/\Delta t$$
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The kinetic energy of the ball increases by $$2mu(u+v)$$
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The kinetic energy of the ball remains the same after the collision
Explanation
In an elastic collision: $$v_{sep}=v_{app}$$
or $$v'-u=v+u$$ or $$v'=v+2u$$
Change in momentum of ball is
$$|p_f-p_i|=|m(-v')-mv|=m(v'+v)=2mu(u+v)$$
Average force is $$\dfrac{\Delta P}{\Delta t}=\dfrac{2m(u+v)}{\Delta t}$$
Change in KE, $$K_f-K_i=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2=2mu(u+v)$$.
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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
When a body falls freely towards the earth, then its total energy:
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Increases
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Decreases
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Remains constant
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First increases and then decreases
Explanation
When a body falls freely towards the earth, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is,
$$potential\ energy + kinetic\ energy = constant$$
It obeys the law of conservation of energy. Due to this, its total energy remains constant.
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
$$W_{2}=KE_{2f}-KE_{2i}=\dfrac{1}{2}m(2v)^{2}-\dfrac{1}{2}mv^{2}$$
$$=\dfrac{1}{2}m(4v^{2}-v^{2})=3\left ( \dfrac{1}{2}mv^{2} \right )=3W_{1}$$
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 $$\displaystyle W=\int F.dx=\int F.\dfrac{dx}{dt}.dt$$
$$\displaystyle W= \int^{ln\,2}_{ln\,1}ma\left(\dfrac{dx}{dt}\right)dt$$ $$m=2kg$$
$$\displaystyle =\int^{ln\,2}_{ln\,1}2\times 4e^{2t}\times 2e^{2t}dt$$
$$\displaystyle =16\int^{ln\,2}_{ln\,1}e^{4t}dt=\dfrac{16}{4}[e^{4t}]^{ln\,2}_{ln\,1}$$
$$=\dfrac{16}{4}[e^{4\,ln\,2}-e^{4\,ln\,1}]$$
$$=4[e^{ln\,(2^4)}-e^{ln\,(1^4)}]$$
$$=4[2^4-1^4]=4[15]=60\,Joules$$
A block of mass $$m$$ lies on the top of a rough inclined plane as shown, find the minimum velocity to be given to block, so that it just stops at the bottom of the inclined.
(given $$m=1\,kg$$ $$g=10\,m/s^2,u=0.80$$)
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$$1\,m/s$$
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$$4\,m/s$$
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$$3\,m/s$$
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$$2\,m/s$$
Explanation
Refer Fig. 1
$$N=mg\cos \theta$$
$$f_s=\mu N$$
$$=\mu\,mg\cos(\theta)$$
Initial energy of block $$=E_i$$
$$Ei=\dfrac{1}{2}mv^2+mgh$$
Final energy of block $$=E_f$$
$$E_f=0+0=0$$
KE PE
$$\Delta E=E_f-E_i$$ = Work done by friction
$$=0-\left(\dfrac{1}{2}mv^2+mgh\right)=-\mu mgcos\theta \times l$$
Refer Fig. 2
putting values
$$-\left[\dfrac{1}{2}\times (1) \times v^2\ + \ (1)\times 10\times 3\right]$$
$$= - 0.80\times (1)\times 10\times \dfrac{4}{5}\times 5$$
$$\Rightarrow -30-\dfrac{v^2}{2}=-32$$
$$\Rightarrow 2=\dfrac{v^2}{2}\Rightarrow v^2=4\Rightarrow \boxed{v=2\,m/s}$$
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
$$KE_2=\dfrac{1}{2}(2m)v^2=2(\dfrac{1}{2}mv^2)=2K$$
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