CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 11 - MCQExams.com

The work done by the force $$\vec{F}=A(y^{2}\hat{i}+2x^{2}\hat{j})$$, where $$A$$ is a constant and $$x$$ & $$y$$ are in meters around the path shown is :

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  • $$zero$$
  • $$Ad$$
  • $$Ad^{2}$$
  • $$Ad^{3}$$
A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is
  • $$u + v$$ away from the wall
  • $$2u + v$$ away from the wall
  • $$|u - v|$$ away from the wall
  • $$|v- 2u |$$ away from the wall
A force $$\vec{F}=(3t\hat{i}+5\hat{j})$$ N acts on a particle whose position vector varies as $$\vec{S}=(2t^{2}\hat{i}+5\hat{j})$$ m, where $$t$$ is time in seconds. The work done by this force from $$t=0$$ to $$t=2s$$ is:
  • 23 J
  • 32 J
  • zero
  • can't be obtained
A particle of mass $$m_0,$$ travelling at speed $$v_0,$$ strikes a stationary particle of mass $$2m_0.$$ As a result, the particle of mass $$m_0$$ is deflected through $$45^o$$ and has a final speed of $$\dfrac{v_0}{\sqrt2}$$. Then the speed of the particle of mass $$2m_0$$ after this collision is 
  • $$\dfrac{v_0}{2}$$
  • $$\dfrac{v_0}{2\sqrt2}$$
  • $$\sqrt2v_0$$
  • $$\dfrac{v_0}{\sqrt2}$$
A particle of mass $$m$$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is $$a$$. A force $$\vec{F}=y\hat{i}-x\hat{j}$$ newton acts on the particle, where $$x, y$$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A ($$a, 0$$) to point B ($$0, a$$) along the circular path is

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  • $$\dfrac{\pi a^{2}}{4}J$$
  • $$\dfrac{\pi a^{2}}{2}J$$
  • $$\dfrac{\pi a^{2}}{3}J$$
  • $$None\ of\ these$$
An elastic string carrying a body of mass $$m$$ at one end extends by $$1.5\ cm$$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:
  • $$3.0\ cm$$
  • $$4.5\ cm$$
  • $$1.5\ cm$$
  • $$9.0\ cm$$
A force $$\vec F=(3t\hat i+5\hat j)$$N acts on a body due to which its displacement varies as $$\vec S=(2t^2\hat i-5\hat j)m$$. Work done by this force in $$2$$ second is
  • 32 J
  • 24 J
  • 46 J
  • 20 J
The world class Moses Mabhida football stadium situated in Durban has a symmetrical arc of length 350 m and a height of 106 m, as shown in the picture below on the left.
The picture on the right shows a funicular (Skycar), which takes tourists to the top of the arc. Suppose that the Skycar with tourists inside, starts from the base of the arc and travels a distance of 175 m along the arc to the viewing platform at the top. Assume that the work done by friction during the Skycars complete ascent is $$5.8\times 10^5J$$. If the combined mass of the Skycar and tourists is 5000 kg, then the work done by the motor that lifts the Skycar is approximately equal to,
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  • $$4.6\times 10^6 J$$
  • $$5.8\times 10^6 J$$
  • $$8.0\times 10^6 J$$
  • $$9.2\times 10^6 J$$
One end of a light spring of spring constant $$k$$ is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is $$+(\dfrac {1}{2})kx^2$$. The possible cases are:
  • The spring was initially compressed by a distance $$x$$ and was finally in its natural length.
  • It was initially stretched by a distance $$x$$ and finally was in its natural length.
  • It was initially in its natural length and finally in a compressed position.
  • It was initially in its natural length and finally in a stretched position.
A spring of natural length $$l$$ is compressed vertically downward against the floor so that its compressed length becomes $$\dfrac {l}{2}$$. On releasing, the spring attains its natural length. If $$k$$ is the stiffness constant of spring, then the work done by the spring on the floor is:
  • Zero
  • $$\dfrac {1}{2}kl^2$$
  • $$\dfrac {1}{2}l(\dfrac {1}{2})^2$$
  • $$kl^2$$
A block of mass $$10kg$$ is released on a fixed wedge inside a cart which is moved with constant velocity $$10 \ ms^{-1}$$ towards right. There is no relative motion between block and cart. Then work done by normal reaction on block in two seconds from ground frame will be $$(g=10 \ ms^{-2})$$:
117433.jpg
  • $$1320 J$$
  • $$960 J$$
  • $$1200 J$$
  • $$240 J$$
A particle of mass m, moving with velocity v collides a stationary particles of mass 2m. As a result of collision, the particle of mass m deviates by $$45^o$$ and has final speed of $$\dfrac {v}{2}$$. For this situation mark out the correct statement (s).
  • The angle of divergence between particles after collision is $$\dfrac {\pi}{2}$$
  • The angle of divergence between particles after collision is less than $$\dfrac {\pi}{2}$$
  • Collision is elastic
  • Collision is inelastic
A spring of spring constant $$5\times 10^3$$ N/m is stretched initially by $$5$$ cm from the unstretched position. The work required to further stretch the spring by another $$5$$ cm is:
  • 6.25 N-m
  • 12.50 N-m
  • 18.75 N-m
  • 25.00 N-m
The force exerted by a compression device is given by $$ F(x) = kx (x-l)$$  for $$ 0 \leq  x \leq  l$$, where $$l$$ is the maximum possible compression, $$x$$ is the compression  and $$k$$ is the constant. Work done to compress the device by a distance $$d$$ will be maximum when 
  • $$d = \dfrac {l}{4}$$
  • $$d = \dfrac {l}{\sqrt{2}}$$
  • $$d = \dfrac {l}{2}$$
  • $$d = l $$
A spring of force constant $$k$$ is cut in two parts at its one-third length. When both the parts have same elongation, the work done in the two parts will be (Spring constant of a spring is inversely proportional to length of spring)
  • Equal to both
  • Greater for the longer part
  • Greater for the shorter part
  • Data insufficient
The relationship between the force $$F$$ and position $$x$$ of a body is as shown in figure. The work done by force $$F$$, in displacing the body from $$x=1 \ m$$ to $$x=5 \ m$$ will be
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  • 30 J
  • 15 J
  • 25 J
  • 20 J
A pendulum was kept horizontal and released. Find the acceleration of the pendulum when it makes an angle $$ \theta $$ with the vertical.

134428_d53edfb0b74b403ebc2b1ccca7cfe9c9.png
  • $$ g\ \sqrt { 1+3\cos ^{ 2 }{ \theta } } $$
  • $$ g\ \sqrt { 1+3\sin ^{ 2 }{ \theta } } $$
  • $$ g\ \sin { \theta } $$
  • $$ 2g\ \cos { \theta } $$
A mass $${ m }_{ 1 }$$ with initial speed $${ v }_{ 0 }$$ in the positive $$x$$-direction collides with a mass $${ m }_{ 2 }=2{ m }_{ 1 }$$ which is initially at rest at the origin, as shown in figure. After the collision $${ m }_{ 1 }$$ moves off with speed $${ v }_{ 1 }={ v }_{ 0 }/2$$ in the negative $$y$$- direction, and $${ m }_{ 2 }$$ moves off with speed $${ v }_{ 2 }$$ at angle $$\theta$$. Find the velocity (magnitude and direction) of the centre of mass after the collision :

133472_e5427870fd9c44bb8219995507be0d06.png
  • $${ v }_{ 0 }/3$$
  • $${ v }_{ 0 }/2$$
  • $${ v }_{ 0 }/5$$
  • $${ v }_{ 0 }/6$$
A body of mass $$m$$ is hauled from the Earth's surface by applying a force $$\vec{F}$$ varying with the height of ascent $$y$$ as $$\vec{F}=2(ay-1)mg$$, where $$a$$ is a positive constant. Find the work performed by this force $$W$$ and the increment in the body's potential energy $$\Delta U$$ in the gravitational field of the Earth over the first half of the ascent.
  • $$\displaystyle W=\frac{3mg}{4a}$$, $$\displaystyle\Delta U=\frac{mg}{2a}$$
  • $$\displaystyle W=\frac{3mg}{a}$$, $$\displaystyle\Delta U=\frac{mg}{2a}$$
  • $$\displaystyle W=\frac{3mg}{4a}$$, $$\displaystyle\Delta U=\frac{mg}{a}$$
  • $$\displaystyle W=\frac{3mg}{4a}$$, $$\displaystyle\Delta U=\frac{mg}{3a}$$
A particle P of mass m attached to vertical axis by two strings AP and BP of length l each. The separation AB=l. The point p rotates around the axis with an angular velocity $$\omega $$. the tension in two strings are $$ { T }_{ 1 }$$ and $$ { T }_{ 2 } $$
taut only if $$ \omega >\sqrt { \frac { 2g }{ l }  } $$

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  • $$ { T }_{ 1 }$$= $$ { T }_{ 2 }$$
  • $$ { T }_{ 1 }$$+ $$ { T }_{ 2 }$$=$$\cfrac{2}{\sqrt3}m\omega^2l $$
  • $$ { T }_{ 1 }$$- $$ { T }_{ 2 }$$=2mg
  • BP will remain
One end of a light spring of spring constant $$k$$ is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is $$+\left(\displaystyle\frac{1}{2}\right){kx^2}$$.The possible cases are:
  • The spring was initally compressed by a distance $$x$$ and was finally in its nautral length
  • It was initially streteched by a distance $$x$$ and finally was in its natural length
  • It was initially in its natural length and finally in compressed position
  • It was initially in its natural length and finally in a stretched position
A spring of constant $$k$$ is fixed to a wall. A boy stretches this spring by distance $$x$$ and in the mean time the compartment moves by a distance $$s$$. The work done by boy with reference to earth is
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  • $$\displaystyle \dfrac{1}{2} kx^2$$
  • $$\displaystyle \dfrac{1}{2} (kx)(s +x)$$
  • $$\displaystyle \dfrac{1}{2} kxs$$
  • $$\displaystyle \dfrac{1}{2} kx(s + x + s)$$
One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is $$1/2 \:k\:x^2$$. The possible cases are
  • the spring was initially compressed by a distance x, was finally in its natural length
  • it was initially stretched by a distance x and was finally in its natural length
  • it was initially in its natural length and finally in a compressed position
  • It was initially in its natural length and finally in the stretched position
Two blocks A and B of masses m and 2m respectively placed on a smooth floor are connected by a spring. A third body C of mass m moves with velocity $$v_0$$ along the line joining A and B and collides elastically with A. At a certain instant of time after collision it is found that the instantaneous velocities of A and B are same then :
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  • the common velocity of A and B at time t$$_0$$ is v/3.
  • the spring constant is k $$= \displaystyle \frac{3mv_0^2}{2x_0^2}$$.
  • the spring constant is k $$= \displaystyle \frac{2mv_0^2}{3x_0^2}$$.
  • none of these
A brick of mass $$1.8$$ kg is kept on a spring of spring constant $$K = 490  N  m^{-1}$$. The spring is compressed so that after the release brick rises to $$3.6$$ m. Find the compression in the spring. (Take $$g= 10 \ m/s^2$$)
  • 0.21 m
  • 0.322 m
  • 0.414 m
  • 0.514 m
A force $$F = - K (y  \widehat i + x  \widehat j)$$ (where $$K$$ is positive constant) acts on a particle moving in the $$x-y$$ plane. Starting from the origin, the particle is taken along the $$x$$-axis to the point $$(a, 0)$$ and then parallel to $$y$$-axis to the point $$(0, a)$$. The total work done by the force $$F$$ on the particle is:
  • $$- 2 Ka^2$$
  • $$2 Ka^2$$
  • $$- Ka^2$$
  • $$Ka^2$$
The work done by the spring in case (i):
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  • $$\displaystyle \frac{kl^2}{2} - \frac{rl^3}{3}$$
  • $$\displaystyle \frac{kl^2}{2} + \frac{rl^2}{3}$$
  • $$\displaystyle \frac{kl^3}{6} - \frac{rl^2}{8}$$
  • $$\displaystyle \frac{kl^3}{8} - \frac{rl^2}{6}$$
A force $$F = - K (y  \widehat i + x  \widehat j)$$, (where $$K$$ is a +ve constant) acts on a particle moving in xy plane starting from origin, the particle is taken along the positive x-axis to the point ($$a, 0$$) and then parallel to y axis to the point ($$a, a$$). The total work done by force $$F$$ on the particle is
  • $$- 2 Ka^2$$
  • $$2 Ka^2$$
  • $$- Ka^2$$
  • $$ Ka^2$$
The work done by the tension T in the above process is

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  • $$Zero$$
  • $$T(L-L\cos \theta )$$
  • $$-TL$$
  • $$-TL\,\sin \theta $$
Two springs have their force constant as $$k_1$$ and $$k_2\:(k_1 > k_2)$$. When they are stretched by the same force
  • no work is done in case of both the springs.
  • equal work is done in case of both the springs
  • more work is done in case of second spring
  • more work is done in case of first spring.
A stone is tied to a string of length, $$\ell$$, 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
  • $$\sqrt{2g\ell}$$
  • $$\sqrt{2(u^2-g\ell)}$$
  • $$\sqrt{u^2-g\ell}$$
  • $$u-\sqrt{u^2-2g\ell}$$
Calculate the work done on the tool by $$\vec{F}$$ if this displacement is along the straight line $$y =x$$ that connects these two points.
  • $$2.50 J$$
  • $$500 J$$
  • $$50.6 J$$
  • $$2 J$$
A ball attached to one end of a string swings in a vertical plane such that its acceleration at point A (extreme position) is equal to its acceleration at point B (mean position). The angle $$\displaystyle \theta $$ is
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  • $$\displaystyle \cos ^{-1}\left ( \frac{2}{5} \right )$$
  • $$\displaystyle \cos ^{-1}\left ( \frac{4}{5} \right )$$
  • $$\displaystyle \cos ^{-1}\left ( \frac{3}{5} \right )$$
  • None of these
A force $$\displaystyle F= -\frac{k}{x^{2}}\left ( x\neq 0 \right )$$ acts on a particle in x-direction. Find the work done by this force in displacing the particle from. $$\displaystyle x= +a\:$$to$$\:x= +2a.$$ Here, $$k$$ is a positive constant.
  • $$\displaystyle  -\frac{k}{2a}$$
  • $$\displaystyle  \frac{k}{2a}$$
  • $$\displaystyle  -\frac{k}{a}$$
  • $$\displaystyle  +\frac{k}{a}$$
An object is displaced from position vector $$\displaystyle \vec{r}_{1}= \left ( 2\hat{i}+3\hat{j} \right )m\:$$ to$$ \:\vec{r}_{2}= \left ( 4\hat{i}+6\hat{j} \right )$$ m under a force $$\displaystyle \vec{F}= \left ( 3x^{2}\hat{i}+2y\hat{j} \right )N.$$ Find the work done by this force.
  • $$83 J$$
  • $$41.5 J$$
  • $$166 J$$
  • $$164 J$$
Which of the following statements is correct regarding the work done $$\vec{F}$$ along these two paths.
  • Work done on x-axis is zero
  • Work done on x-axis is less than on y-axis
  • Work done on x-axis is more than on y-axis but not zero
  • Data insufficient
A simple pendulum is released from rest with the string in horizontal position. The vertical component of the velocity of the bob becomes maximum, when the string makes an angle $$\displaystyle \theta $$ with the vertical. The angle $$\displaystyle \theta $$ is equal to
  • $$\displaystyle \frac{\pi }{4}$$
  • $$\displaystyle \cos ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$
  • $$\displaystyle \sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$
  • $$\displaystyle \frac{\pi }{3}$$
The work done by the varying force in changing the angular displacement from 0 to $$\theta $$ is

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  • $$Wh$$
  • $$FL\,\sin \theta$$
  • $$Fh$$
  • $$\dfrac{1}{2}FL\,\sin \theta $$
A particle is moving in a circular path in the vertical plane. It is attached at one end of a string of length $$l$$ whose other end is fixed. The velocity at lowest point is $$u$$. The tension in the string is $$\displaystyle \vec{T}$$ and acceleration of the particle is $$\displaystyle \vec{a}$$ at any position. Then $$\displaystyle \vec{T}.\vec{a}$$ is zero at highest point if
  • $$\displaystyle u> \sqrt{5gl}$$
  • $$\displaystyle u= \sqrt{5gl}$$
  • Both (a) and (b) correct
  • Both (a) and (b) are wrong
A cutting tool under microprocessor control has several forces acting on it. One force is $$\vec{F}=-\alpha xy^2\hat{j}$$, a force in the negative y-direction whose magnitude depend on the position of the tool. The constant is $$\alpha  = 2.50\ N/m^3$$. Consider the displacement of the tool from the origin to the point  $$x = 3.00 \,m, y = 3.00 \,m$$. Calculate the work done on the tool by $$\vec{F}$$ if the tool is first moved out along the x-axis to the point $$x = 3.00m, \:y= 0m $$ and then moved parallel to the y-axis to $$x = 3.00m, y = 3.00 \,m$$.
  • $$67.5 J$$
  • $$85 J$$
  • $$102 J$$
  • $$7.5 J$$
A particle of mass $$m$$ is suspended by a string of length $$l$$ from a fixed rigid support. A sufficient horizontal velocity $$\displaystyle v_{0}= \sqrt{3gl}$$ is imparted to it suddenly. Calculate the angle made by the string with the vertical when the acceleration of the particle is inclined to the string by $$\displaystyle 45^{\circ}.$$
  • $$\displaystyle \theta = \frac{\pi }{2}$$
  • $$\displaystyle \theta = \frac{\pi }{3}$$
  • $$\displaystyle \theta = \frac{\pi }{4}$$
  • $$\displaystyle \theta = \pi $$
A small spherical ball is suspended through a string of length l. The whole arrangement is placed in a vehicle which is moving with velocity v. Now, suddenly the vehicle stops and ball starts moving along a circular path. If tension in the string at the highest point is twice the weight of the ball then (Assume that the ball completes the vertical circle)
  • $$\displaystyle v= \sqrt{5gl}$$
  • $$\displaystyle v= \sqrt{7gl}$$
  • velocity of the ball at highest point is $$\displaystyle \sqrt{gl}$$
  • velocity of the ball at the highest point is $$\displaystyle \sqrt{3gl}$$
just after it comes in contact with the peg.
241648_4ace1cfb86d1451eb9c94bb16a1b0f31.png
  • $$\displaystyle \frac{mg}{2}$$
  • $$\displaystyle mg$$
  • $$\displaystyle \frac{3mg}{2}$$
  • $$\displaystyle \frac{5mg}{2}$$
A ball tied to the end of the string swings in a vertical circle under the influence of gravity.
  • When the string makes an angle $$\displaystyle 90^{\circ}$$ with the vertical, the tangential acceleration is zero and radial acceleration is somewhere between minimum and maximum.
  • When the string makes an angle $$\displaystyle 90^{\circ}$$ with the vertical, the tangential acceleration is maximum and radial acceleration is somewhere between maximum and minimum.
  • At no place in circular motion, tangential acceleration is equal to radial acceleration.
  • When radial acceleration has its maximum value, the tangential acceleration is zero.
just before the sphere comes in contact with the peg.
241642_960f8fb22b3b47b08916aad356b6539f.png
  • $$\displaystyle \frac{mg}{2}$$
  • $$\displaystyle {mg}$$
  • $$\displaystyle \frac{3mg}{2}$$
  • $$\displaystyle \frac{5mg}{2}$$
if AB is a massless rod,
241618_a6dcbbca76ca462fa697f86d5b4ad63f.png
  • $$\displaystyle \frac{L}{2}$$
  • $$\displaystyle \frac{3L}{2}$$
  • $$\displaystyle \frac{5L}{2}$$
  • $$\displaystyle \frac{7L}{2}$$
The simple $$2 kg$$ pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at $$B$$ and continues in the smaller are in the vertical plane. Calculate the magnitude of the force $$R$$ supported by the pin at $$B$$ when the pendulum passes the position $$\displaystyle \theta = 30^{\circ}.\left ( g= 9.8m/s^{2} \right )$$
241754_c466f74d21ca402b97c3f122452090c2.png
  • $$15 N$$
  • $$30 N$$
  • $$45 N$$
  • $$60 N$$
The bob of the pendulum shown in figure describes an arc of circle in a vertical plane. If the tension in the cord is $$2.5\ times$$ the weight of the bob for the position shown. Find the velocity and the acceleration of the bob in that position.
242842_c97ec45661eb4d77ad78708ce148128f.png
  • $$\displaystyle 16.75ms^{-1}, 5.66ms^{-2}$$
  • $$\displaystyle 5.66ms^{-1}, 16.75ms^{-2}$$
  • $$\displaystyle 2.88ms^{-1}, 16.75ms^{-2}$$
  • $$\displaystyle 5.66ms^{-1}, 8.34ms^{-2}$$
The mass of the bob of a simple pendulum of length $$L$$ is $$m$$. If the bob is left from its horizontal position then the speedof the bob and the tension in the thread in the lowest position of the bob will be respectively:
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  • $$\;\sqrt{2gL}\;and\;3\;mg$$
  • $$\;3\;mg\;and \sqrt{2gL}$$
  • $$\;2\;mg\;and\;\sqrt{2gL}$$
  • $$\;2\;gL\;and\;3\;mg$$
The sphere at A is given a downward velocity $$\displaystyle v_{0}$$ of magnitude $$5 m/s$$ and swings in a vertical plane at the end of a rope of length $$l=2 m$$ attached to a support at $$O$$. Determine the angle $$\displaystyle \theta $$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.
242846_5dfc491e997f4e928267e07c4150a5ea.png
  • $$\displaystyle \sin ^{-1}\left ( \frac{1}{4} \right )$$
  • $$\displaystyle \sin ^{-1}\left ( \frac{1}{3} \right )$$
  • $$\displaystyle \sin ^{-1}\left ( \frac{1}{2} \right )$$
  • $$\displaystyle \sin ^{-1}\left ( \frac{3}{4} \right )$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers