1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 32 InterviewSolution

1551.	A curved suface is shown in figure. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C. Wioth the surface AB, ball 1 has large enough friction to cause rolling down without slipping, ball 2 has a small friction and ball 3 has a negligible friction. (a) For which ball is total mechanical energy conserved? (b) Which ball(s) can reach D? (c )For ball which do not reach D, which of the balls can reach back A?
Answer» (a) As ball 1 rolls down `AB` without slipping, friction is large enough to cause only rotation motion. However, no energy is lost due to friction. Ball 3 has negligible friction. Therefore, total mechanical energy is conserved in case of balls 1 and 3. Small friction in case of ball 2 involves some dissipation of energy. (b) Ball 3 alone can cross over `C` (lower than `A` ) and reach `D` as it has negligible friction. Ball 1 acquires some rotational energy and ball 2 loses some energy overcoming friction. They cannot reach `D`. (c) Balls 1 and 2 turn back before reaching `C`. Ball 2 cannot reach back to `A` on account of frictional losses. Ball 1 cannot roll back to `A` because of kinetic friction.

Discussion

1552.	On complete combustion , a litre of petrol gives off heat equivalent to `3xx10^(7)J` . In a test drive, a car weighing 1200kg, including the mass of driver, runs 15kg per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive. If the efficiency of the car engine were `0.5.
Answer» Here, 1 litre of petrol gives off `3xx10^(7)J` of energy. As efficiency of engine is `0.5,` energy used by car `E=0.5xx3xx10^(7)J=1.5xx10^(7)J` If F is force of friction , then `Fxxs=E` `Fxx(15xx10^(3))=1.5xx10^(7)` `F=10^(3)N`

Discussion

1553.	The rate of work done is (a) energy(b) force (c) power (d) energy flow
Answer» Correct answer is (c) power

Discussion

1554.	What is the unit of work done?
Answer» Joule is the unit of work done.

Discussion

1555.	The unit of power is (a) J (b) W (c) J s (d) both (b) and (c)
Answer» (d) both (b) and (c)

Discussion

1556.	On complete combustion a litre of petrol gives off heat equivalent to 3 × 107 J. in a test drive a car weighing 1200 kg, including the mass of driver, runs 15 km per litre while moving whit a uniform speed on a straight track. Assuming the friction offered by the road surface and air to be uniform, calculate the force of friction acting on the card during the test drive, if the efficiency of the car engine were 0.5.
Answer» ∴ Energy given by car in 1 litre petrol = 0.5 × 3 × 10⁷ J = 1.5 × 10⁷ Joule With 0.5 efficiency, 1 litre generates 1.5 × 10⁷ J. which is used for 15 km drive Fd = 1.5 × 10⁷ J, where d = 1.5 × 10⁴ m Force of friction F = \(\frac{1.5\times 10^7\,J}{1.5\times 10^4\, m}\) F = 10³ N

1556.

On complete combustion a litre of petrol gives off heat equivalent to 3 × 107 J. in a test drive a car weighing 1200 kg, including the mass of driver, runs 15 km per litre while moving whit a uniform speed on a straight track. Assuming the friction offered by the road surface and air to be uniform, calculate the force of friction acting on the card during the test drive, if the efficiency of the car engine were 0.5.

Answer»

∴ Energy given by car in 1 litre petrol

= 0.5 × 3 × 10⁷ J

= 1.5 × 10⁷ Joule

With 0.5 efficiency, 1 litre generates 1.5 × 10⁷ J. which is used for 15 km drive

Fd = 1.5 × 10⁷ J, where d = 1.5 × 10⁴ m

Force of friction F = \(\frac{1.5\times 10^7\,J}{1.5\times 10^4\, m}\)

F = 10³ N

Discussion

1557.	Define 1 Joule of work.
Answer» When a force of 1 Newton acts on an object and the object moves a distance of 1 metre in the direction of the force, the work done by the force is 1 Joule.

Discussion

1558.	What is the unit of work?(a) Newton(b) Joule(c) Watt(d) None of these
Answer» Correct answer is (b) Joule

Discussion

1559.	Two springs of spring constants `1500N//m` and `3000N//m` respectively are streched by the same force. The potential energy gained by the two springs will be in the ratioA. `4:1`B. `1:4`C. `2:1`D. `1:2`
Answer» Correct Answer - C

Discussion

1560.	A block of mass m is directly pulled up slowly on a smooth inclined plane of height h and inclination `theta` with the help of a string parallel to the incline. Which of the following statement is incorrect for the block when it moves up from the bottom to the top the incline? . .A. Work done by the normal reaction force is zero .B. Work done by the string is `mgh`C. Work done by gravity is `mgh`D. Net work done on the block is zero
Answer» Correct Answer - C (a) `W_(N) =NS cos 90^(@) -0` (b) `W_(T) =TS cos 0^(@) =(mg sin theta) ((h)/(sintheta))` `=mgh` (c ) `W_(mg) =(mg) (h) cos180^(@) =-mgh` (d) `W_("Total") =DeltaK =K_(f)-K_(i) =0`. As block is moved slowly or `K_(f) =K_(i)`.

Discussion

1561.	A block is pulled a distance x along a rough horizontal table by a horizontal string. If the tension in the string is T, the weight of the block is W, the reaction is N and frictonal force is F. Write down expressions for the work done by each of these forces.
Answer» `W_(T)=(T) (x) cos 0^@ =Tx` `W_(W) (x) cos 90^@ =0` `W(N)=(N)(x) cos 90^@=0` `W(F)=(F)(x) cos 180^@ =-Fx`.

Discussion

1562.	A particle of mass `0.2` kg is moving is one dimension under a force that delever is a constant preses `0.5 W` in the particle . If the detail speed `(n ms^(-1))` of the particle is zero , the speed`(mms^(-1))` after `5 xx` is
Answer» Here `Delta K.E. = W = P xx t ` `:. (1)/(2) mnu^(2) = P xx t ` `:. Nu = sqrt (2Pt)/(m) = sqrt(2 xx 0.5 xx 5 )/(0.2)) = 5 ms^(-1) `

Discussion

1563.	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 m`B. at `x= 2m`C. at bottommost point of the planeD. at `x = 2.5 m`
Answer» Correct Answer - D `F("net") =mg sin theta -mu ng cos theta` `=mg sin theta-0.3 xmg cos theta`…. (i) At maximum speed `F_("net") =0`. Because after this net force Eq.(i), `F_("net") =0` at `x=(tantheta)/(0.3) =(3/4)/(0.3) =2.5 m`

Discussion

1564.	A `1.8kg` block is moved at constant speed over a surface for which coefficient of friction `mu=1/4` it is pulled by a force F acting at `45^@` with horizontal as shown in Fig. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity. .
Answer» Correct Answer - A::B::C `N=mg-F sin 45^@=18-F/(sqrt2)` Moving with constant speed means net force=0. `F cos 45^@-muN=1/4(18-F/(sqrt2))` `:. (4F)/(sqrt2)=18-(F)/(sqrt2)` `:. F=(18sqrt2)/(5)N` (a) `W_(F)=FS cos45^@` `=((18sqrt2)/5)(2)(1/(sqrt2))=7.2J` (b) `W_(f)=(muN)(S) cos 180^@` `=(1/4)(18-(F)/(sqrt2))(2) (-2)` (c) `W_(mg)=(mg)(s) cos 90^@ =0`.

Discussion

1565.	A particle is projected from a point at an angle with the horizontal at `t=0`. At an instant t, if p is linear momentum, x is horizontal displacement, y is vertical displacement, and E is kinetic energy of the particle, then which of the following graphs are correct?A. (a) B. (b) C. (c) D. (d)
Answer» Correct Answer - A::B::C::D At any time t, `v=sqrt(u^2+g^2t^2-2ug tsintheta)` `E=1/2mv^2=1/2m(u^2+g^2t^2-2ug t sin theta)` Hence, `E-T` graph is parabolic `E=(p^2)/(2m)` Hence, `E-p^2` graph is straight line through origin `E=1/2m u^2-mgy` Putting `y=xtantheta-(gx^2)/(2u^2cos^2theta)` we get `E=1/2m u^2-mg x tan theta+(mx^2x^2)/(2u^2cos^2theta)` Hence, `E-y` graph is a straight line and `E-x` graph is parabolic.

Discussion

1566.	Given in Fig, are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, Indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Answer» (a) We know that Total energy E = KE + PE, kinetic energy can never be negative. In the region between x = 0 & x = a. Potential energy is ‘0’. So, kinetic energy y is positive. In region x > a the potential energy has a value greater than ‘E’. So kinetic energy will be negative in this region. Hence the particle cannot be present in the region x > a. (b) Here PE > E, the total energy of the object and as such the kinetic energy of the object would be negative. Thus object cannot be present in any region on the graph. (c) Here x = 0 to x = a & x > b, the P E is more then E so, K E is negative. The particle cannot be present in these portions. (d) The object cannot exist in the region between x =\(\frac{-b}{2}\) to x =\(\frac{-a}{2}\) & x =\(\frac{-a}{2}\) to x =\(\frac{-b}{2}\) Because in this region P E > E.

1566.

Given in Fig, are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, Indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer»

(a) We know that Total energy E = KE + PE, kinetic energy can never be negative. In the region between x = 0 & x = a.

Potential energy is ‘0’. So, kinetic energy y is positive. In region x > a the potential energy has a value greater than ‘E’. So kinetic energy will be negative in this region. Hence the particle cannot be present in the region x > a.

(b) Here PE > E, the total energy of the object and as such the kinetic energy of the object would be negative. Thus object cannot be present in any region on the graph.

(c) Here x = 0 to x = a & x > b, the P E is more then E so, K E is negative. The particle cannot be present in these portions.

(d) The object cannot exist in the region between

x =\(\frac{-b}{2}\) to x =\(\frac{-a}{2}\) & x =\(\frac{-a}{2}\) to x =\(\frac{-b}{2}\)

Because in this region P E > E.

Discussion

1567.	When an air bubble rises in water, what happens to its potential energy?
Answer» Potential energy of air bubble decrease and kinetic energy increases, because work is done by Upthrust on the bubble.

Discussion

1568.	What happens to the P.E. of a bubble when it rises in water?
Answer» The P.W. of a bubble Decreases when it rises in water.

Discussion

1569.	A bullet is fired normally on an immovable wooden plank of thickness 2 m. It loses 20% of its kinetic energy in penetrating a thickness 0.2 m of the plank. The distance penetrated by the bullet inside the wooden plank is (a) 0.2 m (b) 0.8 m (c) 1 m (d) 1.5 m
Answer» (c) 1 m The wood offers a constant retardation. If the bullet loses 20% of its kinetic energy by penetrating 0.2m. it can penetrate further into 4 × 0.2 = 0.8 m with the remaining kinetic energy. So the total distance penetrated by the bullet is 0.2 + 0.8 = 1 m.

1569.

A bullet is fired normally on an immovable wooden plank of thickness 2 m. It loses 20% of its kinetic energy in penetrating a thickness 0.2 m of the plank. The distance penetrated by the bullet inside the wooden plank is (a) 0.2 m (b) 0.8 m (c) 1 m (d) 1.5 m

Answer»

(c) 1 m

The wood offers a constant retardation. If the bullet loses 20% of its kinetic energy by penetrating 0.2m. it can penetrate further into 4 × 0.2 = 0.8 m with the remaining kinetic energy. So the total distance penetrated by the bullet is 0.2 + 0.8 = 1 m.

Discussion

1570.	A bullet of mass 0.5 kg slows down from a speed of 5 m s-1 to that of 3 m s-1. Calculate the change in kinetic energy of the ball.
Answer» Mass of ball = 0.5kg Initial velocity = 5m/s Initial kinetic energy = 1/2 × mass x (velocity)² = 1/2 ×0.5 ×(5)² = 1/2 ×0.5 ×25=6.25J Final velocity of the ball = 3m/s Final kinetic energy of the ball =1/2 × mass x (velocity)² = 12 × 0.5 x (3)² = 12 ×0.5 x 9 = 2.25J Change in the kinetic energy of the ball = 2.25 J − 6.25J = − 4J There is a decrease in the kinetic energy of the ball

1570.

A bullet of mass 0.5 kg slows down from a speed of 5 m s-1 to that of 3 m s-1. Calculate the change in kinetic energy of the ball.

Answer»

Mass of ball = 0.5kg

Initial velocity = 5m/s

Initial kinetic energy = 1/2 × mass x (velocity)²

= 1/2 ×0.5 ×(5)²

= 1/2 ×0.5 ×25=6.25J

Final velocity of the ball = 3m/s

Final kinetic energy of the ball =1/2 × mass x (velocity)²

= 12 × 0.5 x (3)²

= 12 ×0.5 x 9 = 2.25J

Change in the kinetic energy of the ball = 2.25 J − 6.25J = − 4J

There is a decrease in the kinetic energy of the ball

Discussion

1571.	When two bodies collide elastic, then:A. Kinetic energy of the system alone is conservedB. Only momentum is conservedC. Both energy and momentum are conservedD. Neither energy nor momentum is conserved
Answer» Correct Answer - C

Discussion

1572.	A body of mass 2 kg is moved from a point A to a point B by an external agent in a conservative force field. If the velocity of the body at the points A and B are `5 m//s` and `3 m//s` respectively and the work done by the external agent is - 10 J, then the change in potential energy between point A and B isA. 6 JB. 36 JC. 16 JD. None of these
Answer» Correct Answer - A

Discussion

1573.	A particle is moved from (0, 0) to (a, a) under a force a `F=(3hat(i)+4hat(j))` from two paths. Path 1 is OP and path 2 is OPQ. Let `W_(1)` and `W_(2)` be the work done by this force in these two paths. Then,A. `W_(1)=W_(2)`B. `W_(1)=2W_(2)`C. `W_(2)=2W_(1)`D. `W_(2)=4W_(1)`
Answer» Correct Answer - A

Discussion

1574.	The diagram respresent the potential energy `U` of a function of the inter atomic distance `r`. Which diagram corresponding to slabe molecules found in nature?A. B. C. D.
Answer» Correct Answer - A When the distance between atoms is large its introtamic force is very weak when they one close force because increases and it a period distance force becomes zero When they are further brought closer force becomes repative in motion This can explated by slope of `U` curve shown in graph

Discussion

1575.	A particle is moved from `(0,0)`to `(a,a)` under a force `bar F = (3hati + 4hatj)` and path `2is OQP`.Let `W_(1) and W_(2)` be the work by done this force in these to paths .Then . A. `W_(1)=W_(2)`B. `W_(1)=2W_(2)`C. `W_(2)=2W_(1)`D. `W_(2)=4W_(1)`
Answer» Correct Answer - A (a) Given force is a constant force and work done by a constant force is always path independent. `:. " " W_(1)=W_(2)`

Discussion

1576.	A boy weighing `50 kg` finished long jump at a distance of `8 m`. Considering that he moved along a parabolic path and his angle of jumps was `45^(@)`, his initial `KE` will be `g = 9.8 m//s^(2)`A. `960 J`B. `1560 J`C. `2460 J`D. `1960 J`
Answer» Correct Answer - B Maximum range`= (u^(2))/(g) = R or u^(2) = Rg` `KE = (1)/(2) m u^(2)= (1)/(2) mRg = (1)/(2) xx 50 xx8 xx 9.8 = 1960 J`

Discussion

1577.	A particle is moved from `(0,0)`to `(a,a)` under a force `bar F = (3hati + 4hatj)` and path `2is OQP`.Let `W_(1) and W_(2)` be the work by done this force in these to paths .Then . A. `W_(1) = W_(2)`B. `W_(1) = 2W_(2)`C. `W_(2) = 2W_(1)`D. `W_(2) =4 W_(1)`
Answer» Correct Answer - A Given force is a constant force and wolt down is per independent inder a constant force Hence `W_(1) - W_(2)`

Discussion

1578.	A stone of mass 1 kg tied to a light inextensible string of lenth `L=(10)/(3)m`, whirling in a circular path in a vertical plane. The ratio of maximum tension to the minimum tension in the string is 4. If g is taken to be `10ms^(-2)`, the speed of the stone at the highest point of the circle isA. `20 m//sec`B. `10 sqrt3 m//sec`C. `5 sqrt2 m//sec`D. `10 m//sec`
Answer» Correct Answer - D Since the maximum tension `T_(B)` in the string moving to the vertical circle is at the bottom and minimum tension `T_(T)` is at the top. `:. T_(B) = (mv_(B)^(2))/(L) + mg and T_(T) = (mv_(T)^(2))/(L) - mg` `:. (T_(B))/(T_(T)) = ((mv_(B)^(2))/(L) + mg)/((mv_(T)^(2))/(L) - mg) = (4)/(1) or (v_(B)^(2) + gL)/(v_(T)^(2) - gL) = (4)/(1)` or `v_(B)^(2) + gL = 4v_(T)^(2) - gL but v_(B)^(2) = v_(T)^(2) + 4gL` `:. v_(T)^(2) + 4gL+ gL = 4v_(T)^(2) - 4gL implies 3v_(T)^(2) = 9gL` `:. v_(T)^(2) = 3 xx gxx L = 3 xx 10 xx (10)/(3) or v_(T) = 10 m//sec`

Discussion

1579.	A particle is dropped a height h. A constant horizontal velocity is given to the particle. Taking g to be constant every where, kinetic energy E of the particle w. r. t. time t is correctly shown inA. B. C. D.
Answer» Correct Answer - A

Discussion

1580.	Assertion : A work done in moving a body over a closed loop is zero for every force in nature. Reason : Work done does not depend on nature of force.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 - D

Discussion

1581.	A bullet is fired fram a riffie . If the rifle recoils freely determine whether the kinetic energy of the rifle is greater then , equal or less then that of the bullet .
Answer» `K.E = (p^(2)/(2m)` For equal of p, `K.E prop (1)/(mass)`

Discussion

1582.	A bullet is fired fram a riffie . If the rifle recoils freely determine whether the kinetic energy of the rifle is greater then , equal or less then that of the bullet .A. Less than that of the bulletB. More than that of the bulletC. Same as that of the bulletD. Equal or less than that of the bullet
Answer» Correct Answer - A

Discussion

Explore topic-wise InterviewSolutions in .

The rate of work done is (a) energy(b) force (c) power (d) energy flow

What is the unit of work done?

The unit of power is (a) J (b) W (c) J s (d) both (b) and (c)

Define 1 Joule of work.

What is the unit of work?(a) Newton(b) Joule(c) Watt(d) None of these

Two springs of spring constants `1500N//m` and `3000N//m` respectively are streched by the same force. The potential energy gained by the two springs will be in the ratioA. `4:1`B. `1:4`C. `2:1`D. `1:2`

A block is pulled a distance x along a rough horizontal table by a horizontal string. If the tension in the string is T, the weight of the block is W, the reaction is N and frictonal force is F. Write down expressions for the work done by each of these forces.

A particle of mass `0.2` kg is moving is one dimension under a force that delever is a constant preses `0.5 W` in the particle . If the detail speed `(n ms^(-1))` of the particle is zero , the speed`(mms^(-1))` after `5 xx` is

A `1.8kg` block is moved at constant speed over a surface for which coefficient of friction `mu=1/4` it is pulled by a force F acting at `45^@` with horizontal as shown in Fig. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity. .

A particle is projected from a point at an angle with the horizontal at `t=0`. At an instant t, if p is linear momentum, x is horizontal displacement, y is vertical displacement, and E is kinetic energy of the particle, then which of the following graphs are correct?A. (a) B. (b) C. (c) D. (d)

When an air bubble rises in water, what happens to its potential energy?

What happens to the P.E. of a bubble when it rises in water?

A bullet is fired normally on an immovable wooden plank of thickness 2 m. It loses 20% of its kinetic energy in penetrating a thickness 0.2 m of the plank. The distance penetrated by the bullet inside the wooden plank is (a) 0.2 m (b) 0.8 m (c) 1 m (d) 1.5 m

A bullet of mass 0.5 kg slows down from a speed of 5 m s-1 to that of 3 m s-1. Calculate the change in kinetic energy of the ball.

When two bodies collide elastic, then:A. Kinetic energy of the system alone is conservedB. Only momentum is conservedC. Both energy and momentum are conservedD. Neither energy nor momentum is conserved

A particle is moved from (0, 0) to (a, a) under a force a `F=(3hat(i)+4hat(j))` from two paths. Path 1 is OP and path 2 is OPQ. Let `W_(1)` and `W_(2)` be the work done by this force in these two paths. Then,A. `W_(1)=W_(2)`B. `W_(1)=2W_(2)`C. `W_(2)=2W_(1)`D. `W_(2)=4W_(1)`

The diagram respresent the potential energy `U` of a function of the inter atomic distance `r`. Which diagram corresponding to slabe molecules found in nature?A. B. C. D.

A particle is moved from `(0,0)`to `(a,a)` under a force `bar F = (3hati + 4hatj)` and path `2is OQP`.Let `W_(1) and W_(2)` be the work by done this force in these to paths .Then . A. `W_(1)=W_(2)`B. `W_(1)=2W_(2)`C. `W_(2)=2W_(1)`D. `W_(2)=4W_(1)`

A boy weighing `50 kg` finished long jump at a distance of `8 m`. Considering that he moved along a parabolic path and his angle of jumps was `45^(@)`, his initial `KE` will be `g = 9.8 m//s^(2)`A. `960 J`B. `1560 J`C. `2460 J`D. `1960 J`

A particle is moved from `(0,0)`to `(a,a)` under a force `bar F = (3hati + 4hatj)` and path `2is OQP`.Let `W_(1) and W_(2)` be the work by done this force in these to paths .Then . A. `W_(1) = W_(2)`B. `W_(1) = 2W_(2)`C. `W_(2) = 2W_(1)`D. `W_(2) =4 W_(1)`

A particle is dropped a height h. A constant horizontal velocity is given to the particle. Taking g to be constant every where, kinetic energy E of the particle w. r. t. time t is correctly shown inA. B. C. D.

A bullet is fired fram a riffie . If the rifle recoils freely determine whether the kinetic energy of the rifle is greater then , equal or less then that of the bullet .