1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 16 InterviewSolution

751.	A body is being raised to a height h from the surface of earth. What is the sign of work done by (a) applied force (b) gravitational force ?A. Positive, PositiveB. Positive, NegativeC. Negative, PositiveD. Negative, Negative
Answer» Correct Answer - B Work done, `W=vecF.vecd=Fd cos theta` where `theta` is the angle between the force `vecF` and displacement `vecd` (i) As the body moves along the direction of applied force. ` therefore theta =0^@`, `W = Fd cos theta= Fd`. It is positive. (ii) As the body moves in a direction opposite to the gravitational force which acts vertically downwards. `therefore theta=180^@`, `W= Fd cos180^@ =-Fd`. It is negative.

Discussion

752.	State carefully if the following quantities are positive or negative: 1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. 2. work done by gravitational force ‘in the above case, 3. work done by friction on a body sliding down an Inclined plane, 4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, 5. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer» 1. We know that: W = \(\vec F,\vec S\) FS cosθ ‘Positive’ Reason: Force is acting in the direction of displacement (θ = 0°) 2. ’Negative’ Reason: Force is acting in the opposite direction to displacement (θ = 180°) 3. ’Negative’ Reason: Force of friction is opposite to the displacement (θ = 180°) 4. ‘Positive’ Reason: The body mover in the direction of force applied (θ = 0°) 5. ‘Negative’ Reason: The resistive force opposes the motion (θ = 0°)

752.

State carefully if the following quantities are positive or negative: 1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. 2. work done by gravitational force ‘in the above case, 3. work done by friction on a body sliding down an Inclined plane, 4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, 5. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer»

1. We know that: W = \(\vec F,\vec S\) FS cosθ

‘Positive’

Reason: Force is acting in the direction of displacement (θ = 0°)

2. ’Negative’

Reason: Force is acting in the opposite direction to displacement (θ = 180°)

3. ’Negative’ Reason: Force of friction is opposite to the displacement (θ = 180°)

4. ‘Positive’ Reason: The body mover in the direction of force applied (θ = 0°)

5. ‘Negative’ Reason: The resistive force opposes the motion (θ = 0°)

Discussion

753.	The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. work done by gravitational force in the above case, work done by friction on a body sliding down an inclined plane, work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer» Positive In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket. Negative In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative. Negative Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case. Positive Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive. Negative The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

753.

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. work done by gravitational force in the above case, work done by friction on a body sliding down an inclined plane, work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer»

Positive

In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

Negative

In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.

Negative

Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.

Positive

Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.

Negative

The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

Discussion

754.	Which of the following potential energy curves in Figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between the centres of balls.
Answer» The potential energy of a system of two masses varies inversely as the distance (r) between them i.e. V(r) < \(\frac{1}{r} \) When the two balls touch each other, PE becomes zero i.e. at r – R + R = 2R , v (r) = 0. Out of the given graphs only the curve (V) satisfies these two conditions.

754.

Which of the following potential energy curves in Figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between the centres of balls.

Answer»

The potential energy of a system of two masses varies inversely as the distance (r) between them i.e.

V(r) < \(\frac{1}{r} \) When the two balls touch each other, PE becomes zero i.e. at r – R + R = 2R , v (r) = 0. Out of the given graphs only the curve (V) satisfies these two conditions.

Discussion

755.	Is friction a conservative force? Give reason.
Answer» No, because work done in a closed path against friction is not zero. Moreover, and the work done against fiction depends on the path.

Discussion

756.	What is the significance of the negative sign in w = – mgh.
Answer» A negative sign signifies that the work is done against the gravitational force.

Discussion

757.	The momentum of a body is increased by 50%. What is the percentage change in its K.E.?
Answer» When the momentum is increased by 50%, velocity increase by 50%, i.e., velocity becomes \(\frac{3}{2}\) times, so, K.E. becomes \(\frac{9}{4}\) times, i.e. \(\frac{9}{4}\) × 100 = 225%. Hence increase in K.E. = 225 – 100 = 125%.

Discussion

758.	Potential energy of a body is the energy possessed by the body by virtue of its position. P.E.`=mgh` where the symbols have their usual meaning. Kinetic energy of a body is the energy possessed by the body by virtual of its velocity. `K.E.=(1)/(2)m upsilon^(2)` Energy can neither be created nor be destroyed. However energy can be changed from one form to other, such that energy appearing in one form is equal to the energy disappearing in the other form. With the help of the passage given above, choose the most appropriate alternative for each of the following question `:` The ratio of potential energy to kinetic energy at a height of `62.5 m` above the ground isA. 2B. 1C. 3D. 4
Answer» Correct Answer - B When `h` is half the original height, `P.E.= mgh` also becomes half. The other half of total energy appears as kinetic energy. Therefore, `PE//KE=1`

Discussion

759.	When is the exchange of energy maximum during an elastic collision?
Answer» Exchange of energy during an elastic collision is maximum `(=100%)` , when the two colliding bodies have equal masses.

Discussion

760.	A potential energy function for a two-dimensional force is the form `U=3x^2y-7x`. Find the force that acts at the point `(x,y)`.
Answer» `F_x=-(delU)/(delx)=-(del(3x^3y-7x))/(delx)=-(9x^2y-7)=7-9x^2y` `F_y=-(delU)/(dely)=-(del(3x^3y-7x))/(dely)=-(3x^3-0)=-3x^3` Thus, the force acting at point `(x, y)` is `vecF=F_xhati+F_yhatj=(7-9x^2y)hati-3x^3hatj`

Discussion

761.	The potential energy function for a particle executing linear SHM is given by `V(x)= 1/2kx^2` where k is the force constant of the oscillator. For `k = 0.5 Nm^(-1)`, the graph of V(x) versus x is shown in the figure A particle of total energy E turns back when it reaches `x=pmx_m`.if V and K indicate the potential energy and kinetic energy respectively of the particle at `x= +x_m`,then which of the following is correct? A. V=O,K=EB. V=E,K=OC. V lt E, K=OD. V=O, K lt E
Answer» Correct Answer - B (b) Total energy is `E=PE+KE " "……….(i)` When particle is at `x=x_(m)` i.e., at extreme position, returns back. Hence, at `x=x_(m)`,x=0,KE=0 `E=PE+0=PE=V(x_(m))=(1)/(2)kx_(m)^(2)`

Discussion

762.	A particle located in a one-dimensional potential field has its potential energy function as `U(x)(a)/(x^4)-(b)/(x^2)`, where a and b are positive constants. The position of equilibrium x corresponds toA. (a) `(b)/(2a)`B. (b) `sqrt((2a)/(b))`C. (c) `sqrt((2b)/(a))`D. (d) `(a)/(2a)`
Answer» Correct Answer - B The position of equilibrium corresponds to `F(x)=0` Since `F(x)=(-dU(x))/(dx)` so `F(x)=-(d)/(dx)(a/x^4-b/x^2)` or `F(x)=(4a)/(x^5)-(2b)/(x^3)` For equilibrium, `F(x)=0`, therefore `(4a)/(x^5)-(2b)/(x^3)=0impliesx=+-sqrt((2a)/(b))` `(d^2U(x))/(dx^2)=-(20a)/(x^6)+(8b)/(x^4)` Putting `x=+-sqrt((2a)/(6))` gives `(d^2U(x))/(dx^2)` as negative So U is maximum. Hence, it is position of unstable equilibrium

Discussion

763.	The potential energy function for a particle executing linear SHM is given by `V(x)= 1/2kx^2` where k is the force constant of the oscillator. For `k = 0.5 Nm^(-1)`, the graph of V(x) versus x is shown in the figure A particle of total energy E turns back when it reaches `x=pmx_m`.if V and K indicate the potential energy and kinetic energy respectively of the particle at `x= +x_m`,then which of the following is correct? A. V=0, K=EB. V=E, K=0C. V lt K, K=0D. V=0, K lt E
Answer» Correct Answer - B At any instant, the total energy of an oscillator is the sum of kinetic energy and potential energy. Total energy, E=K + V At `x= +x_m`, the particle turns back, therefore its velocity at this point is zero, i.e. v=0 `therefore` K=0 `therefore E=1/2kx^2 =V`

Discussion

764.	A smooth block of mass m moves up from bottom to top of a wedge which is moving with an acceleration `a_(10)`. Find the work done by the pseudo force measured by the person sitting at the edge of the wedge.
Answer» The observer moving with wedge will observe a pseudo force `ma_0` in the direction opposite to `veca_0`. ` ` Since `vecF_(ps)=-vecma_0` is a constant force, therefore, `W_(ps)=vecF_(ps)vecS=-(ma_0)lcos180^@=ma_0l`

Discussion

765.	A ball of mass `m` is released from A inside a smooth wedge of mass `m` as shown in figure. What is the speed of the wedge when the ball reaches point B? A. (a) `((gR)/(3sqrt2))^(1//2)`B. (b) `sqrt(2gR)`C. (c) `((5gR)/(2sqrt3))^(1//2)`D. (d) `sqrt(3/2gR)`
Answer» Correct Answer - A Let the velocity of wedge be v. Loss in PE=Gain in KE `mgR cos 45^@=1/2mv^2+1/2m(v_1cos45^@-v)^2+1/2m(v_1sin45^@)^2` From conservation of linear momentum `m(v_1cos45^@-v)=mv` Here `v_1` is the velocity of ball w.r.t. wedge, Solve to get `v=((gR)/(3sqrt2))^(1//2)`

Discussion

766.	What is the source of the kinetic energy of the falling rain drops?
Answer» It is the gravitational potential energy which is being converted into kinetic energy.

Discussion

767.	The potential energy of a particle of mass m is given by `U=(1)/(2)kx^(2)` for `x lt 0` and U = 0 for `x ge 0`. If total mechanical energy of the particle is E. Then its speed at `x = sqrt((2E)/(k))` isA. ZeroB. `sqrt(2E)/m`C. `sqrt(E/m)`D. `sqrt(E/(2m))`
Answer» Correct Answer - B

Discussion

768.	Assertion: The change in kinetic energy of a particle is equal to the work done on it by the net force. Reason: Change in kinetic energy of particle is equal to the work done in case of a system of one particle.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If asserti on is true but reason is falseD. If the assertion and reason both are false
Answer» Correct Answer - C

Discussion

769.	In gravitational field, the work done in moving a body from one point into another depends on (a) initial and final positions (b) distance between them (c) actual distance covered(d) velocity of motion
Answer» (c) initial and final positions

Discussion

770.	If maching is lubricated with oilA. (a) The mechanical advantage of the machine increasesB. (b) The mechanical efficiency of the machine increasesC. (c) Both its mechanical advantage and efficiency increaseD. (d) Its efficiency increases, but its mechanical advantage decreases
Answer» Correct Answer - B Friction decreases due to lubrication, which increases efficiency. Mechanical advantage depends upon geometrical construction of machine.

Discussion

771.	If maching is lubricated with oilA. the machanical adventage of the maching of the maching incresesB. the machanical efficiency of the maching of the maching incresesC. both its machanical advantage and afficiency incresesD. in efficiency increases but its machanical advantage decreases
Answer» Correct Answer - B (b) Mechanical officiency `= (Output work)/(lnput energy)` The output work will increase became the friction becams less . Thus the machanical efficincy increase

Discussion

772.	Block A has no relative motion with resprct to wedge fixed to the lift as shown in figure during motion-1 or motion-2 Then, A. work done by gravity on block A in motion-2 is less then in motion-1B. wrok done by normal reaction on block A in both be motions will be positive.C. work done by force of friction in motion-1 may be positive.D. work done by force of friction in motion-1 may be positive.
Answer» Correct Answer - A::B::C::D (a) Work done by gravity in motion 1 is zero `(theta=90^@)` and in motion 2 is negative `(theta=180^@)` (b) In both cases angle between N and S is acute. (c) and (d) Depnding on the act upnthe plane or down the plane. Therefore angle between friction and displacement may be obtuse ot acute. So, work done by friction may be negtive or positive.

Discussion

773.	A sphere of mass m held at a height `2R` between a wedge of same m and a rigid wall, is released from, Assuming that all the surfaces are frictionless. Find the speed of the bodies when the sphere hits the ground. .
Answer» If speed of sphere is v downwards then speed of wedge at this instant will be v cot `alpha` in horizontal direction. Now, Decrease in potention energy of sphere=increase in kinetic energy of both `:. mgR=1/2 mv^(2) + 1/2 (v cos alpha)^(2)` `=1/2mv^(2) alpha` `:. v=sqrt(2gR) sin alpha=`speed of sphere and speed of wedge `=v cot alpha` `=sqrt(2gR) cos alpha`.

Discussion

774.	Block A has a weight of `3000 N` and block B has a weight of `500 N`. Determine the distance that A must descend from rest before it obtains a speed of `2.5 m//s`. Neglect the mass of the cord and pulleys. .
Answer» If A descends X then B will ascend 2x. Further if speed of A at this instant is `2.5 m//s`, then speed of B at this instant will be `5 m//s`. Now, Deecrease in potential energy of `A =` increase in potential energy of `B +` increase in kinetic energy of both `:. (300)x =(50)(2x) + 1/2 ((300)/(9.8)) (2.5)^(2)` `+(1)/(2) ((50)/(9.8))(5.0)^(2)` Solving we get, `x=0.796 m`.

Discussion

775.	A point particle of mass m, moves long the uniformly rough track PQR as shown in figure. The coefficient of friction, between the particle and the rough track equals `mu`. The particle is released, from rest from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The value of the coefficient of friction `mu` and the distance x `(=QR)`, are, respectively close to: A. `0.2` and `6.5m`B. `0.2` and `3.5m`C. `0.29` and `3.5m`D. `0.29` and `6.5m`
Answer» Correct Answer - C As energy lost over `PQ=` energy lost over `QR` `:. (mu mg cos theta)PQ=(mu mg)QR` or `QR=PQcos theta` .....(i) From figure, `sin theta=sin 30^(@)=(2)/(PQ)=(1)/(2)` `PQ=4m` From (i), `QR=4 cos 30^(@)=4 (sqrt(3))/(2)=2sqrt(3)m=3.5m` Again, decrease in `P.E.=` loss of energy due to friction in `PQ` and `QR` `mgh=(mu mg cos theta) PQ+mu mg xx QR` `h=mu cos thetaxxPQ+mu mg xx QR` `h=mu cos thetaxx PQ +mu xx QR` `2= mu cos 30^(@)xx4+muxx2sqrt(3)` `=mu(4xx(sqrt(3))/(2)+2sqrt(3))=muxx4sqrt(3)` `mu=(2)/(4sqrt(3))=(1)/(2sqrt(3))=0.29`

Discussion

776.	In a thermal station, coal is used for the generation of electricity. Mention how energyh changes from one form to the other before it is trnsformed into electrical energy?
Answer» When coal is burnt, heat energy is produced. This converts water into steam. This steam rotates the tubine and thus heat energy is converted into mechanical energy of rotation. The generator converts this meachnical energy into electrical energy.

Discussion

777.	The work done by a fielder when he takes a catch in a cricket match, is negative Explain.
Answer» Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.

Discussion

778.

Difference between power and horse power.

Answer»

Power	Horsepower
Power is defined as the work done per unit time or the rate at which energy is consumed.	A horsepower is a unit of measurement of power used to express the power of engines or cars.
power = work/time SI unit of power is watts.	Horsepower is itself a bigger unit of power.
1 watt = 1 joule/sec	1 horsepower = 746 watt

Discussion

779.

Differentiate between Energy and power.

Answer»

ENERGY	POWER
The capability to do work is called energy.	Power is defined as the work done per unit time or the rate at which energy is consumed.
SI unit of energy is Joule. 1 Joule = 1 Newton× 1 meter	SI unit of power is watt. 1 watt = 1 joule/sec
Energy can change from one form to another.	Power cannot be changed from one form to another.
Energy can be stored	Power cannot be stored.

Discussion

780.

Differentiate between Potential Energy and Kinetic Energy.

Answer»

POTENTIAL ENERGY	KINETIC ENERGY
The energy of a body due to its higher position above the earth or due to change in the shape of the body is called potential energy.	The energy of a body due to its motion or velocity is called kinetic energy.
Potential energy = m×g×h	Kinetic energy = \(\frac{1}{2}\)mv²
for example, a brick lying on the roof of a house at a height possess potential energy.	for example, a running motorcycle has kinetic energy
Potential energy cannot be transferred between objects.	Kinetic energy can be transferred between objects.

Discussion

781.

Differentiate between work and power.

Answer»

Work	Power
Work is said to be done when a force produces motion.	Power is defined as the work done per unit time or the rate at which energy is consumed.
SI unit of work is joule.	SI unit of power is watt.
Work done = Force × dispacement	Power = Work/ Time
Work done is independent of time taken.	power of an object or body depends on the time taken.

Discussion

782.	Illustrate the law of conservation of energy by discussing the free fall of a body.
Answer» According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another. whenever energy changes from one form to another, the total amount of energy remains constant. The Law of conservation of energy is valid in all situations and for all kinds of energy transformations. For example: When a body is at some height from earth it possesses only potential energy, as the body falls downwards then its potential energy goes on decreasing but kinetic energy goes on increasing. Just before hitting the ground entire potential energy is transformed into kinetic energy and it possesses only kinetic energy. Total energy= kinetic + potential energy At maximum height, kinetic energy = 0 Total energy = potential energy On hitting ground, potential energy = 0 Total energy = kinetic energy

782.

Illustrate the law of conservation of energy by discussing the free fall of a body.

Answer»

According to the law of conservation of energy,

energy can neither be created nor be destroyed.

It can only be transformed from one form to another. whenever energy changes from one form to another, the total amount of energy remains constant.

The Law of conservation of energy is valid in all situations and for all kinds of energy transformations.

For example: When a body is at some height from earth it possesses only potential energy, as the body falls downwards then its potential energy goes on decreasing but kinetic energy goes on increasing.

Just before hitting the ground entire potential energy is transformed into kinetic energy and it possesses only kinetic energy.

Total energy= kinetic + potential energy

At maximum height, kinetic energy = 0

Total energy = potential energy

On hitting ground, potential energy = 0

Total energy = kinetic energy

Discussion

783.	Water is falling on the block of a turbine a height of `25 m xx 3 xx 10^(3) kg` of water pours on the blade per minute. If the whole of energy is transforred to the turbine, power delivered is `g = 9.8 m//s^(2)`A. `12250 W`B. `16250 W`C. `8250 W`D. `20250 W`
Answer» Correct Answer - A Power = Energy // time = mgh//h.

Discussion

784.	A particle of mass m moving towards west with speed v collides with another particle of mass m movies towards south. If two particles st ich t o each other the speed of the new particle of mass 2 m will beA. `vsqrt2`B. `(v)/(sqrt2)`C. v/2D. v
Answer» Correct Answer - B

Discussion

785.	In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of `2 m s^(-1)` at `45^@` from height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be `10 m s^(-2)`, the kinetic energy of the shotput when it just reaches the ground will beA. 2.5 JB. 5.0 JC. 52.5 JD. 155.0 J
Answer» Correct Answer - D Initial KE of the shotput `=1/2(10 kg) (1 m//s^2)=5 J` Initial PE of the shotput at a height 1.5 m above ground `=mgh = (10 kg) (10 m//s^2) (1.5 m)=150 J` Total initial energy of the shotput = 155 J Since air resistance is negligible and final potential of the shotput as it hits the ground (h= 0) is zero, kinetic energy of the shotput on hitting the ground =155 J

Discussion

786.	An unruly demonstrator lift a stone of mass `m` kg from the ground and throws it at this opponent. At the time of projection, the stone is at height `h` meter above the ground and has a speed of `v(m)//(s)` what horse power does he use?A. `(1)/(2)(mv^2)/(746)`B. `(mgh)/(746)`C. `((1)/(2)mv^2+mgh)/(746)`D. `((1)/(2)mv^2-mgh)/(746)`
Answer» Correct Answer - C `hp=(mgh+(1)/(2)mv^2)/(746)`

Discussion

787.	A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to `(-K//r^(2))`, where k is a constant. The total energy of the particle is -A. `-k/r`B. `-k/(2r)`C. `k/(2r)`D. `(2k)/r`
Answer» Correct Answer - B Since the particle is moving in horizontal circle, centripetal force, `F=(mv^2)/r=k/r^2,mv^2=k/r` ...(i) Kinetic energy of the particle, `K=1/2mv^2 =k/(2r)` (Using (i)) As `F=(-dU)/(dr)` `therefore` Potential energy, `U= - underset(oo)oversetrintFdr=-undersetoooversetrint((-k)/r^2)dr =kundersetoooversetrintr^(-2) dr=(-k)/r` `therefore` Total energy = K+U = `k/(2r)-k/r =(-k)/(2r)`

Discussion

788.	If a man increase his speed by 2m/s, his K.E. is doubled, the original speed of the man is
Answer» Man possesses kinetic energy, because of its velocity (v). If m is mass of man, then `K=(1)/(2)mv^(2)` Given, `" " v_(1)=v, m_(1)=m_(2)=m`, When `" " v_(2)=(v+2)ms^(-1)` Then `" " K_(2)=2K_(1)` `:.(K_(1))/(K_(2))=(v_(1)^(2))/(v_(2)^(2))implies(K_(1))/(2K_(1))=(v^(2))/(v+2)^(2)impliesv^(2)-4v-4=0` This gives, `v_(1)=(4+sqrt((16+16)))/(2)=(4+sqrt(32))/(2)` `v_(1)=2(sqrt2+1)ms^(-1)`

Discussion

789.	A particle of mass m has half the kinetic energy of another particle of mass `m/2`. If the speed of the heavier particle is increased by `2 ms^(-1)` its new kinetic energy becomes equal to t he original kinetic energy of the lighter particle. The ratio of the orighinal speeds of the lighter and heavier particle isA. `1:1`B. `1:2`C. `1:3`D. `1:4`
Answer» Correct Answer - B

Discussion

790.	Which of the diagram shown in figures respresents variation of total mechanical energy of a pendulam oscillation in air as function of time?A. B. C. D.
Answer» (c ) When a pendulum oscillates in air, it will lose energy continously in overcoming resistance due to air. Therefore total mechanical energy of the pendulum decreases continously wity time. The variation is correctly represented by curve(c ).

Discussion

791.	Constant as eliptical rail `PQ` in the varticle plain with `OP = = 3m ` and `OQ= 4 m` . A block of mass `1` kg is pailed along the rail from `P` to `Q` with a force of `18 N` , which is always parallel to less `PQ` Assuming are frictionless losess , the kinetic energy the block when `0` reches `Q` is `( nxx 10)` pales . THe velie of a (Take acceleration due to gravity ) `= 10ms^(-2))`
Answer» work done = increase in potential energy +gain in kinetic energy `F xx d = mgh + gain in K. E. ` `18 xx 5 = 1 xx 10 xx 4 + gin in K. E. ` `:. Gain in K.E. = 50 J =10 n` `:. N = 5 ` .

Discussion

792.	A charged particle X moves directly towards another charged particle Y. For the X plus Y system, the total momentum is p and the total energy is E.A. (a) p and E are conserved if both X and Y are free to move.B. (b) (a) is true only if X and Y have similar charges.C. (c) If Y is fixed, E is conserved but not P.D. (d) If Y is fixed, neither E nor P is conserved.
Answer» Correct Answer - A::C Energy will be conserved whether Y is fixed or free to move because no dissipating force is present. But momentum will be conserved only if Y is free to move, because then external force will be zero.

Discussion

793.	A particle is projected with a velocity `u` making an angle `theta` with the horizontal. The instantaneous power of the gravitational forceA. (a) Varies linearly with timeB. (b) Is constant throughoutC. (c) Is negative for complete pathD. (d) None of the above
Answer» Correct Answer - A At any time, `vecv=[u cos theta]hati+(u cos theta-g t)hatj` `P=vecF.vecv=(-mghatj)*[u cos thetahatj+(u sin theta-g t)hatj]` `=mg^2t-mgu sin theta` Hence, power varies linearly with time.

Discussion

794.	A particle is projected vertically upwards with a speed of `16ms^-1`. After some time, when it again passes through the point of projection, its speed is found to be `8ms^-1`. It is known that the work done by air resistance is same during upward and downward motion. Then the maximum height attained by the particle is (take `g=10ms^-2`)A. (a) `8m`B. (b) `4.8m`C. (c) `17.6m`D. (d) `12.8m`
Answer» Correct Answer - A From work-energy theorem, for upward motion `1/2m(16)^2=mgh+W` (work done by air resistance) for downward motion, `1/2m(8)^2=mgh-Wimplies1/2[(16)^2+(8)^2]=2gh` or `h=8m`

Discussion

795.	A uniform force of `(3 hat(i)+hat(j))` newton acts on a particle of mass `2 kg`. Hence the particle is displaced from position `(2 hat(i)+hat(j))` meter to position `(4 hat(i)+ 3hat(j)-hat(k))` meter. The work done by the force on the particle is `:`A. ` 15 J`B. `9 J`C. `6 J`D. `13 J`
Answer» Correct Answer - B Here, `vec(F)=(3hat(i)+hat(j))N` `vec(r)=vec(r_(2))-vec(r_(1))=(4hat(i) +3 hat (j) -k)-(2hat(i)+hat(k))` `=(2hat(i)+3hat(j)-2hat(k))` `W=vec(F).vec(r)=(3hat(i)+hat(j)).(2hat(i)+3hat(j)-2hat(k))` `W=6+3=9J`

Discussion

796.	The potential energy of a particle of mass 1 kg `U = 10 + (x-2)^(2)`. Herer, U is in joule and x in met. On the positive x=axis particle travels up to x=+6cm. Choose the wrong statement.A. On negative X-axis particle travels upto x=-2 mB. The maximum kinetic energy of the particle is 16 JC. Both (a) and (b) are correctD. Both (a) and (b) are incorrect
Answer» Correct Answer - C (c ) At x=6 m, U=26J (extreme position) On the other side `U=26=10+(x-2)^(2)" or "x-2=-+4` `:." "`x=6m and x=-2m `U_(min)=10J, "at " x=2` `KE_(max)=E-U_(min)=16J, "at "x=2`

Discussion

797.	A force of F=0.5 N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, g=10 `ms^(-2)`) A. `-0.5 J`B. 0.5 JC. 2 JD. `-2 J`
Answer» Correct Answer - B (b) Maximum acceleration of 1 kg block may be `a_(max)=mug=1ms^(-2)` Common acceleration, without relative motion between two blocks may be, `a=(0.5)/(3)ms^(-2)` Since, `a lt a_(max)` There will be no relative motion and blocks will move with acceleration `(0.5)/(3)ms^(-2)`. Force of friction by lower block on upper block, `f=ma=(1)((0.5)/(3))=(1)/(6)N("towards right")` `:. " " W=fxxs=0.5J`

Discussion

798.	A body is moving is down an inclined plane of slope `37^@` the coefficient of friction between the body and the plane varies as `mu=0.3x`, where x is the distance traveled down the plane by the body. The body will have maximum speed. `(sin 37^@ =(3)/(5))` .A. At x=1.16 mB. At x=2 mC. At bottom of planeD. At x=2.5 m
Answer» Correct Answer - D (d) Body will have maximum speed where, `mg sin theta=mu mg cos theta` or `" " sin37^(@)=(0.3x).cos37^(@)" or " x=2.5 m`

Discussion

799.	A block of mass `2.5 kg` is pushed `2.20`m along a frictionless horizontal table by a constant force 16 N directed `45^@` above the horizontal. Determine the work done by. (a) the applied force, (b) the mormal force exerted by the table, (c) the force of gravity and (d) determine the total work done on the block.
Answer» (a) `W_(F) =FS cos 45^@` `=(16)(2.2)((1)/(sqrt(2)))=24.9J` (b) `W=_(N) =N S cos90^@=0` (c) `W_(mg) =(mg)(S) cos 90^@=0` (d) Only three forces are acting. So, total work done is summation fo all above work done.

Discussion

800.	Which of the diagram shown in figures respresents variation of total mechanical energy of a pendulam oscillation in air as function of time?A. B. C. D.
Answer» Correct Answer - C (c ) When a pendulum oscillates in air, it will lose energy continuously in overcoming resistance due to air. Therefore, total mechanical energy of the pendulum decreases continously with time. The variation is correctly represented by curve (c ).

Discussion

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