1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 18 InterviewSolution

851.	A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is 2 m. If due to air resistance loss of energy is 10%, what is the speed with which the bob arrived at the lowest point.
Answer» \(\frac{1}{2}\)mv² = 90% of mgh ∴ v = 6 m/s

851.

A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is 2 m. If due to air resistance loss of energy is 10%, what is the speed with which the bob arrived at the lowest point.

Answer»

\(\frac{1}{2}\)mv² = 90% of mgh

∴ v = 6 m/s

852.	In a nuclear reactor, a neutron of high speed `(~~10^(7)ms^(-1))` must be slowed down to `10^(3)ms^(-1)` so that it can have a high probality of interacting with isotipe `_92U^(235)` and causing it to fission. Show that a neutron can lose most of its K.E. in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a fewe times the neutron mass. The material making up the light nuclei usually heavy water `(D_(2)O)` or graphite is called modertaor.
Answer» Initial K.E. of neutron is `K_(1)=(1)/(2)m_(1)u_(1)^(2)` Velocity of neutron after collision with deuterium, `upsilon_(1)=((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))` Final K.E. of neutron is `K_(2)=(1)/(2)m_(1)upsilon_(1)^(2)=(1)/(2)m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)u_(1)^(2)` `:.` Frational K.E. retained by neutron is `f_(1)=(K_(2))/(K_(1))=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)` For deuterium, `m_(2)=2m_(1) :. f_(1)=(1)/(9)`, Fractional K.E. lost by neutron `f_(2)=1-f_(1)=1-(1)/(9)=(8)/(9)=90%` This is the fractional K.E. gained by moderating nuclei. Therefore, almost `90% ` of neutron energy is transferred to deuterium. Similarly, in case of carbon, we can show that `f_(1)=28.4%` and `f_(2)=71.6%`.

Discussion

853.	Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature? Why?
Answer» No. W.D. is zero only in case of a conservative force.

Discussion

854.	Give an example in which a force does work on a body but fails to change its K.E.
Answer» When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.

Discussion

855.	How high must a body be lifted to gain an amount of potential energy equal to the kinetic energy it has. When moving at sped 20ms-1. The value of acceleration due to gravity at that place is g = 9.8 ms-2.
Answer» mgh = \(\frac{1}{2}\)mv² i.e. 9.8 h = \(\frac{1}{2}\) × 20 × 20 h = \(\frac{200}{9.8}\)= 20.4 m.

855.

How high must a body be lifted to gain an amount of potential energy equal to the kinetic energy it has. When moving at sped 20ms-1. The value of acceleration due to gravity at that place is g = 9.8 ms-2.

Answer»

mgh = \(\frac{1}{2}\)mv²

i.e. 9.8 h = \(\frac{1}{2}\) × 20 × 20

h = \(\frac{200}{9.8}\)= 20.4 m.

Discussion

856.	The sign of work done by a force on a body is important to understand. State carefullyif the following quantities are positive or negative: work done by a man in lifting abucket out of a well by means of a rope tied to the bucket.work done by gravitational force in the above case, work doneby friction on a body sliding down an inclined plane,work done by an applied force on a body moving on a rough horizontal plane withuniform velocity, work done by the resistive force of air on a vibrating pendulumin 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 offrictional 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.

856.

The sign of work done by a force on a body is important to understand. State carefullyif the following quantities are positive or negative: work done by a man in lifting abucket out of a well by means of a rope tied to the bucket.work done by gravitational force in the above case, work doneby friction on a body sliding down an inclined plane,work done by an applied force on a body moving on a rough horizontal plane withuniform velocity, work done by the resistive force of air on a vibrating pendulumin 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 offrictional 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

857.	Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.
Answer» (i), (ii), (iii), (iv), and (vi) The potential energy of a system of two masses is inversely proportional to the separation between them. In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero (i.e., V(r) = 0) when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

Discussion

858.	Consider the decay of a free neutron at rest: n → p+ e–Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (Fig. 6.19).[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like e–, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e–+ ν]
Answer» The decay process of free neutron at rest is given as: n->p+e^- From Einstein’s mass-energy relation, we have the energy of electron as Δmc² Where, Δm = Mass defect = Mass of neutron – (Mass of proton + Mass of electron) c = Speed of light Δm and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the β-decay of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly explains the continuous energy distribution.

858.

Consider the decay of a free neutron at rest: n → p+ e–Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (Fig. 6.19).[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like e–, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e–+ ν]

Answer»

The decay process of free neutron at rest is given as:

n->p+e^-

From Einstein’s mass-energy relation, we have the energy of electron as Δmc²
Where,

Δm = Mass defect = Mass of neutron – (Mass of proton + Mass of electron)
c = Speed of light

Δm and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the β-decay of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly explains the continuous energy distribution.

Discussion

859.	A body of mass 5 kg moving with a velocity of 10 m/s collides with another body of mass 20 kg at, rest and comes to rest. The velocity of the second body due to collision is A. 2.5 m/sB. 5 m/sC. 7.5 m/sD. 10 m/s
Answer» A. 2.5 m/s Explanations: Mass of 1 body = 5 kg Velocity of 1 body = 10 m/s Mass of 2 body = 20 kg Velocity of 2 body=? Velocity of 2 = (mass of 1) x (velocity of 1)/mass of 2 Velocity of 2 = (5) x (10)/20 Velocity of 2= 2.5 m/s

859.

A body of mass 5 kg moving with a velocity of 10 m/s collides with another body of mass 20 kg at, rest and comes to rest. The velocity of the second body due to collision is A. 2.5 m/sB. 5 m/sC. 7.5 m/sD. 10 m/s

Answer»

A. 2.5 m/s

Explanations:

Mass of 1 body = 5 kg

Velocity of 1 body = 10 m/s

Mass of 2 body = 20 kg

Velocity of 2 body=?

Velocity of 2 = (mass of 1) x (velocity of 1)/mass of 2

Velocity of 2 = (5) x (10)/20

Velocity of 2= 2.5 m/s

Discussion

860.	What is the amount of work done by 1. a weight lifter in holding a weight of 10 kg on his shoulder for 30 s, and 2. a locomotive against gravity, if it is traveling on a level plane?
Answer» 1. zero 2. zero

860.

What is the amount of work done by 1. a weight lifter in holding a weight of 10 kg on his shoulder for 30 s, and 2. a locomotive against gravity, if it is traveling on a level plane?

Answer»

1. zero

2. zero

Discussion

861.	Ram and Ali have been fast friends since childhood. Ali neglected studies and now has no means to earn money other than a channel whereas Ram has become an engineer. Now both are working in the same factory. Ali uses camel to transport the load within the factory. Due to low salary and degradation in health of camel, Ali becomes worried and meets his friend Ram and discusses his problems. Ram collected some data and with some assumptions concluded the following: i. The load used in each trip is `1000kg` and has friction coefficient `mu_k=0.1` and `mu_s=0.2`. ii. Mass of camel is `500kg`. iii. Load is accelerated for first `50m` with constant acceleration, then it is pulled at a constant speed of `5ms^-1` for `2km` and at last stopped with constant retardation in `50m`. iv. From biological data, the rate of consumption of energy of camel can be expressed as `P=18xx10^3v+10^4Js^-1` where P is the power and v is the velocity of the camel. After calculations on different issues, Ram suggested proper food, speed of camel, etc. to his friend. For the welfare of Ali, Ram wrote a letter to the management to increase his salary. (Assuming that the camel exerts a horizontal force on the load): The total energy consumed by the camel during the trip of `2100m` isA. (a) `2.1xx10^6J`B. (b) `4.22xx10^7J`C. (c) `2.22xx10^4J`D. (d) `4.22xx10^6J`
Answer» Correct Answer - B The time of travel in accelerated motion = time of travel in retarded motion. `t_(AB)=t_(CD)=v/a=(5)/(0.25)=20s` Now time for uniform motion `=T_(BC)=2000//5=400s` Total energy consumed is `underset0overset440int Pdt= underset0overset20 int [18xx10^3v+10^4]dt+underset20overset420int[18xx10^3xx5+10^4]dt+underset420overset440int[18xx10^3v+10^4]dt` Putting `vdt=dx` and changing limits approximately, we get `vdt=dx` It becomes `underset0overset50int18xx10^3dx+[10^4t]_0^20+10^5[420-20]+underset(2050)overset(2100)int 18xx10^3dx+[10^4]_(420)^(440)` `=18xx10^3xx50+10^4[20]+10^5xx400+18xx10^3[50]+10^4[20]J` `=4.22xx10^7J`

Discussion

862.	As athlete in the olympic ganes cover a distance of `100 m` in `10 s`. Hiskinetic energy can be estrimated to be in the rangeA. ` 200 J - 500 J`B. `2 xx 10^(5) J - 3 xx 10^(5) J`C. `20000 J - 50000 J`D. `20000 J - 50000 J`
Answer» Correct Answer - D (d) The average speed of the athelete `nu= (100)/(10) = 10 m//s :. K.E = (1)/(2) m nu^(2)` If mass is `40 kg` then , K.E `= (1)/(2) xx 40 xx (10)^(2) = 20000 J ` If is `100 kg` then , K.E `= (1)/(2) xx 100 xx (10)^(2) = 50000 J`

Discussion

863.	An object of mass m has a speed `v_(0)` as it passes throgh the origin. Origin. It subjected to a retaeding force given by `F(x) =-Ax`. Here, A is a positive constant. Find its x-coordinate when it stops.
Answer» Correct Answer - A `K_(f)-K_(i) =W_(F)=intFdx` `0-1/2mv_(0)^(2) =int_(0)^(x)-Axdx` `:. 1/2mv_(0)^(2)=(Ax^(2))/ or x=v_(0)sqrt((m)/(A))`

Discussion

864.	The kinetic energy of a projectile at its highest position is `K`. If the range of the projectile is four times the height of the projectile, then the initial kinetic energy of the projectile is .A. `sqrt2K`B. `2 K`C. `4 K`D. `2sqrt2 K`
Answer» Correct Answer - B At highest point, `(1)/(2)m u_(x)^(2)=K` `:. u_(x)=sqrt((2K)/m)` `R=4 H` `(2u_(x)u_(y))/g=(4u_(y)^(2))/(2g)` `:. u_(y)=u_(x) =sqrt((2K)/m)` Now, `K_(i)1/2m u^(2) =1/2m(u_(x)^(2)+u_(y)^(2))` `=1/2m((2K)/m+(2K)/m)` `=2K`.

Discussion

865.	A projectile is fired from the origin with a velocity `v_(0)` at an angle `theta` with x-axis. The speed of the projectile at an altitude `h` is .A. `v_(0) cos theta`B. `sqrt(v_(0)^(2) - 2gh)`C. `sqrt(v_(0)^(2) sin^(2)theta-2gh)`D. None of these
Answer» Correct Answer - B `K_(i) + U_(i) =K_(f) + U_(f)` `:. 0 + 6=(1)/(2) xx 1 xx v^(2) + 2` `:. v =sqrt(v_(0)^(2)-2gh`.

Discussion

866.	A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. `250 J`B. `450 J`C. `275 J`D. `475 J`
Answer» When a force acts upon a moving body , then the kinekic energy of the body increase and the increase is equal to the work done. This is work energy therom. Work done `= (1)/(2) mv^(2) - (1)/(2) mu^(2) = K_(f) - K_(i)` Another definition of work is force xx displacement. `:. Fdx = K_(f) - (1)/(2) xx mv_(1)^(2)` where the subscripts `f` and l` stand for final and initial. `F. dx = K_(f) - (1)/(2) xx10 xx (10)^(2)` `implies F. dx = K_(f) - 500` `rArr underset(x=20)overset(x=30)int=K_(f)-500` Using the formula `int x^(n) dx = (x^(n+1))/(n+1)`, we have `- 0.1 [(x^(2))/(2)]_(x=20)^(x=30) = K_(f) - 500` `- 0.1 [(30)^(2)/(2) - ((20)^(2))/(2)] = K_(f) - 500` `implies K_(f) - 500 = - 25` `implies K_(f) - 500 - 25 = 475J`

Discussion

867.	A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. 250 JB. 275 JC. 450 JD. 475 J
Answer» Correct Answer - D Here, m=10 kg, `v_i= 10 m s^(-1)` Initial kinetic energy of the block is `K_i=1/2mv_i^2=1/2xx(10 kg)xx(10 m s^(-1))^2 =500 J` Work done by retarding force `W=intF_rdx=underset(20)overset(30)int -0.1 xdx = -0.1 [x^2/2]_20^30` `= -0.1[(900-400)/2]= -25 J` According to work energy theorem `W=K_f-K_i` `K_f=W+K_i = - 25 J + 500 J = 475 J `

Discussion

868.	A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. `475 J`B. `450 J`C. `275 J`D. `250 J`
Answer» Correct Answer - A `K_(f)-K_(i) =W =int Fdx` `:. K_(f) =K_(i) + int_(20)^(30)(-0.1x)dx` `1/2 xx 10 xx (10)^(2)-[0.1(x^(2))/(2)]_(20)^(30)` `475 J`.

Discussion

869.	A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .
Answer» According to work-energy theorem, work done =change in kinetic energy of the vehicle `:. W=K_(f)-K_(i) orF.dx=K_(f)-(1)/(2)mv_(i)^(2)` or `F.dx=k_(f)-(1)/(2)xx10xx(10)^(2)or F.dx =k_(f)-500` or `int_(x=20)^(x=30)(-0.1)x dx=k_(f)-500` or `-0.1[(x^(2))/(2)]_(x=20)^(x=30)=K_(f)-500` or `-0.1[((30)^(2))/(2)-((20)^(2))/(2)]=K_(f)-500` or `K_(f)-500=-0.1(450-200)` or `K_(f)-500=-25` `:. K_(f)=500-25=475 J`

Discussion

870.	A man of mass `m`, standing at the bottom of the staircase of height `L` climbs it and stands at its top.A. Work done by all forces on man is equal to the rise in potential energy `mgL`.B. Work done by all forces on man is zeroC. Work done by the gravitational force on man is `mgL`D. The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
Answer» Correct Answer - B::D When a man of mass `m` climbs up the staircase of height `L`, work done by the gravitation force on man is `(-)mgL`, and work done by muscular force is `mgL`. If we ignore air resistance and friction, then the work done by all forces on man is equal to `-mgL+mgL=zero. `Further , reaction force from a step does not do work because the point of application of force does not move while the force exists.

Discussion

871.	(a) Two masses one n times heavier than the other are dropped from same height. How do their momentum compare just before they hit the ground ? (b) Two masses one n times heavy as the other heave equal knetic energy. How do their momentum compare ?
Answer» (a) When dropped from same height, the two masses hit the ground with same velocity `=sqrt(2gh)` `:. (P_(2))/(P_(1))=(m_(2))/(m_(1))=n` (b) We know that K.E., `K=(p^(2))/(2m)` `:. P=sqrt(2mK)` As K is same, `(p_(2))/(p_(1))=sqrt((m_(2))/(m_(1)))=sqrt(n)`

Discussion

872.	Two constant forces `vecF_1` and `vecF_2` act on a body of mass `8kg`. These forces displace the body from point `P(1, -2, 3)` to `Q` `(2, 3, 7)` in `2s` starting from rest. Force `vecF_1` is of magnitude `9N` and is acting along vector `(2hati-2hatj+hatk)`. Work done by the force `vecF_2` isA. (a) `80J`B. (b) `-80J`C. (c) `-180J`D. (d) `180J`
Answer» Correct Answer - D `vecs=hati+5hatj+4hatk=1/2((vecF_1+vecF_2)/(m))(2)^2` (i) `vecF_1=(9(2i-2j+k))/(sqrt(2^2+2^2+1))=6i-6j+3k` (ii) From Eqs. (i) and (ii), `vecF_2=-2hati+26hatj+13hatk` Work done by `vecF_2: W=vecF_2.vecs=180J`

Discussion

873.	A ball is thrown vertically downwards from a height of 20m with an intial velocity `v_(0)`. It collides with the ground, loses 50% of its energy in collision and rebounds to the same height. The intial velocity `v_(0)` is (Take, g =10 `ms^(-2)`)A. `14ms^(-1)`B. `20ms^(-1)`C. `28ms^(-1)`D. `10ms^(-1)`
Answer» Correct Answer - b (b) Suppose a ball rebounds with speed v, `v=sqrt(2gh)=sqrt(2xx10xx20)` `=20m//s` Energy of a ball just after rebound, `E=(1)/(2)mv^(2)=200m` As, 50% of energy loses in collision means just before collision energy is 400 m. According to law of conservation of energy, we have `(1)/(2)mv_(0)^(2)+mgh=400m` `implies (1)/(2)mv_(0)^(2)+mxx10xx20=400m` `implies v_(0)=20m//s`

Discussion

874.	If two balls each of mass `0.06 kg` moving in opposite directions with speed `4 m//s` collide and rebound with the same speed , then the impulse imparted to each ball due to other isA. `0.49=8kg-m//s`B. `0.24kg-m//s`C. `0.81kg-m//s`D. zero
Answer» Correct Answer - A

Discussion

875.	Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted intoA. Heat energyB. Electrical energyC. Nuclear energyD. Mechanical energy
Answer» Correct Answer - A

Discussion

876.	A man `M_(1)` of mass `80 kg` runs up a staircase in `15 s`. Another man `M_(2)` also of mass `80 kg` runs up the same staircase in `20 s`. The ratio of the power development by then will be:A. (a) `1`B. (b) `4/3`C. (c) `16/9`D. (d) None of the above
Answer» Correct Answer - B Power P=Work/Time Work done by both will be same. Hence, `P_1/P_2=t_2/t_1=20/15=4/3`

Discussion

877.	A `90-kg` man runs up an escalator while it is not in operation in `10s`. What is the average power developed by the man. Suppose the escalator is running so that the escalator steps move at a speed of `0.5m^-1`. What is then the power developed by the man as seen by the ground reference if he moves at the same speed relative to the escalator steps as he did when the escalator is not in operation?
Answer» Work done by the man in climbing the escalator, `W=mgh=90xx10xx20sin30^@=9000J` With the escalator stationary, man takes `10s` to climb up, therefore average power `P=W/t=9000/10=900W` Speed of the man when escalator is stationary, `v=20/10=2ms^-1` In the second case, the speed of the man relative to the moving escalator is also `2ms^-1`. His speed relative to the ground is `0.5+2=2.5ms^-1`. According, the time taken by the man in climbing is `20/2.5=8s` The power developed by the man as seen from the ground reference is then, on the average: `Power=W/l=9000/8=1125W`

Discussion

878.	A small bell starts moving from `A` over `a` fixed track as shown in the figure . Surface `AB` is friction from `A` to `B` the bell roll ralis without sipping `BC` is friction , `K_(A)K_(B)` and` K_(C)` are kinetic energy of the bell at `A, B` and `C` respacecvely.Then A. `h_(A) gt h_(C) , K_(B) gt k_(c)`B. `h_(A) gt h_(C) , K_(c) gt k_(A)`C. `h _(A) = h_(C) , K_(B) gt k_(c)`D. `h_(A) lt h_(C) , K_(B) gt k_(c)`
Answer» Correct Answer - A::B At point `A` , potential energy of the ball `= mgh_(A)` At point `C` , potential energy of the ball `= mgh_(C)` Total energy at point `A,E_(A) = K_(A) + mgh_(A)` Total energy at point `B,E_(A) = K_(A) + mgh_(A)` Total energy at point `C,E_(C) = K_(C) + mgh_(C)` According to the law of conserveation of energy `E_(A) = E_(B) = E_(C) ` ....(i) E_(A) = E_(B) rArr E_(C)gt k_(C)` ....(ii) E_(A) = E_(C)` K_(A) + mgh_(A) = K_(B) + mgh_(C) ` ...(iii) `rArr b_(A) gt b_(C) , K_(C) gt K_(A) ` ...(iv) Option (b) is correct force (i) (ii) and (iv) , we get h_(A) ,K_(B) gt K_(C)` Option (a) is correct .

Discussion

879.	A cylinder of 10 kg is sliding in a plane with an initial velocity of 10 m / s . If the coefficient of friction between the surface and cylinder is 0.5 then before stopping, it will cover. `(g=10m//s^(2))`A. 12.5 mB. 5 mC. 7.5 mD. 10 m
Answer» Correct Answer - D

Discussion

880.	Work done in raising a box depends onA. How fast it is raisedB. The strength of the manC. The height by which it is raisedD. None of the above
Answer» Correct Answer - C

Discussion

881.	A horizontal force of 5 N is required to maintain a velocity of 2 m / s for a block of 10 kg mass sliding over a rough surface. The work done by this force in one minute isA. 600 JB. 60 JC. 6 JD. 6000 J
Answer» Correct Answer - A

Discussion

882.	A light body and a heavy body have same linear momentum. Which one has greater K.E ?
Answer» Lighter body has more K.E. as K.E. = p²/2m and for constant p, K.E. ∝ 1/m.

Discussion

883.	A block is moved from rest through a distance at `4 m` along a string line path . The mass of the block is `5 kg` and the force acting on it is `20N`. If the kinetic energy acquired by the block be `40 J`, at what angle to the path is the force acting?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. None of these
Answer» Let required angle is `theta` Work done = change in `KE` ` implies Fs cos theta = 40 - 0` `implies 20 xx 40 cos theta - 400 implies theta = 60^(@)`

Discussion

884.	A body of mass m was slowly pulled up the hill by a force F which at each point was directed along the tangent of the trajectory. All surfaces aresmooth. Find the work performed by this force. A. mglB. `-mgl`C. mghD. zero
Answer» Correct Answer - C (c ) `W_(F)+W_(g)=DeltaK=0, W_(F)=-W_(g)=mgh`

Discussion

885.	In an aircraft carrier warship the runway is a 20 m long strip inclined at `theta = 20^(@)` to the horizontal. The launcher is effectively a large spring that pushes an aircraft of mass m = 2000 kg for first 5 m of the 20 m long runway. The jet engine of the plane produces a constant thrust of `6 xx 10^(4) N` for the entire length of the runway. The plane needs to have a speed of 180 kph at the end of the runway. Neglect air resistance and calculate the spring constant of the launcher. [`sin 20^(@) = 0.3` and `g = 10 m//s^(2)`]
Answer» Correct Answer - `k=2.096xx10^(5) N//m`

Discussion

886.	An elevator has mass of 600 kg, not including passengers The elevator is designed to ascend at constant speed a vertical distance of 20 m in 15 sec. It is driven by a motor that can provide up to 30 hp to the elevator, what is the maximum number of passengers that can ride in the elevator ? Assume that an average passenger has a mass of 65 kg
Answer» Correct Answer - A

Discussion

887.	The potential energy of the system is represented in the first figure. The force acting on the system will be represented by A. B. C. D.
Answer» Correct Answer - C As slop of position graph is position and constant upto cartain distance and then it becomes zero As from `F = (-dU)/(dx)` up to distance `a, F` = constant (negative and becomes zero suddenly

Discussion

888.	A block of mass M is placed on a horizontal surface having coefficient of friction m. A constant pulling force `F=(Mg)/2` is applied on the block to displace it horizontally through a distance d. Find the maximum possible kinetic energy acquired by the block.
Answer» Correct Answer - `Mgd[(sqrt(u^(2)+1)/2-mu]`

Discussion

889.	A small block is made to slide, starting from rest, along two equally rough circular surfaces from A to B through path 1 and 2. The two paths have equal radii. The speed of the block at the end of the slide was found to be `V_1` and `V_2` for path 1 and 2 respectively. Which one is larger `V_1` or `V_2`?
Answer» Correct Answer - `V_(1)`

Discussion

890.	A ball is thrown vertically upwards with a velocity of 10 `ms^(-1)`. It returns to the ground with a velocity of 9 `ms^(-1)`. If g=9.8 `ms^(-2)` , then the maximum height attained by the ball is nearly (assume air resistance to be uniform)A. 5.1 mB. 4.1 mC. 4.61 mD. 5 m
Answer» Correct Answer - C (c ) Loss of kinetic energy `=k_(i)-K_(f)` `=(1)/(2)m(100-81)=(19)/(2)m " "(m=mass)` Half of the loss will be in upward journey. Hence, `mgh+(1)/(2)((19)/(2)m)=K_(i)=(1)/(2)m(100)=50m` `:. " " h=4.61 m`

Discussion

891.	A ball of mass m is thrown upward with a velocity `upsilon`. If air exerts an average resisting force F, the velocity with which the ball returns to the thrower isA. `sqrt((mg)/(mg+F))`B. `vsqrt((F)/(mg+F))`C. `vsqrt((mg-F)/(mg+F))`D. `vsqrt((mg+F)/(mg-F))`
Answer» Correct Answer - C

Discussion

892.	A block of mass `m` is connected to a spring of spring constant k as shown in figure. The frame in which the block is placed is given an acceleration a towards left. Neglect friction between the block and the frame walls. The maximum velocity of the block relative to the frame is A. (a) `sqrt(m/k)`B. (b) `alphasqrt(m/k)`C. (c) `alphasqrt((m)/(2k))`D. (d) `2alphasqrt(m/k)`
Answer» Correct Answer - B Solving this question relative to the frame (car) of reference. For maximum velocity (relative to frame), the block must be in equilibrium position. Let `x_0` be the equilibrium elongation in spring, then `ma=kx_0` From work-energy theorem, `(mv^2)/(2)-0=-(kx_0^2)/(2)+ma x_0` Solving the above equation, we get `v=asqrt(m/l)` This question is an application of using the work-energy theorem in non-inertial frame of reference.

Discussion

893.	Figure shows the variation of potential energy of a particle as a function of x, the x-coordinate of the region. It has been assumed that potential energy depends only on x. For all other values of x, U is zero, i.e., `xlarr10` and `xgt15`, `U=0`. If the total mechanical energy of the particle is `-40J`, then it can be found in regionA. (a) `xlarr10` and `xgt15`B. (b) `-10ltxlarr5` and `6ltxlt15`C. (c) `10ltxlt15`D. (d) It is not possible
Answer» Correct Answer - D The concept is same as above. If `E=-40J`, `K=-40-U`. It means `Ularr40` for K to be positive, but from given variations, it is clear that minimum, value of `U` is `-35J`. So this is not possible.

Discussion

894.	Figure shows a plot of the potential energy as a function of `x` for a particle moving along the x-axis. Which of the following statement(s) is/are true? A. (a) a, c, and d are points of equilibriumB. (b) a is a point of stable equilibriumC. (c) b is a unstable equilibrium pointD. (d) All of the above
Answer» Correct Answer - D For equilibrium, potential energy has to be minimum, maximum or constant.

Discussion

895.	The graph shows the potential energy of a system of two ions as a function of their interionic separtion. The system of the two icns has a total mechanical energy of `-0.5xx10^(-17)j.` What is the range of possible values for the distance between the two ions? At what separation do the ions have minimum kinetic energy? What is the maximum possible kinetic energy of the ions? what energy is required to breake apart this ionic molecule?
Answer» (i) `1.7xx10^(-10)` m to `5.0xx10^(-10)m` (Outside this range the potential energy would need to be greater than the total energy, that is the kinetic energy would have to be negative which is an impossible situation) (ii) `1.7xx10^(-10)mor5.0xx10^(-10)m(K.E.=0)` (iii) `1.5xx10^(-7)J(at d=3.0xx10^(-10)m)` (iv) `0.5xx10^(-17)J.` (The total mechanical energy must be increased to 0)

Discussion

896.	Figure shows a plot of the potential energy as a function of `x` for a particle moving along the x-axis. Which of the following statement(s) is/are true? A. The particle is in equlibrium at 3 out of 5B. The particle is in equlibrium at 2 ou of 5 points shownC. The particle is in stable equlibrium at `x=b`D. The particle is in unstable equlibrium at `x=d`
Answer» Correct Answer - A::C::D

Discussion

897.	Potential energy U(r ) varies with position r as shown in figure Which of the following conclusions are correct?A. Force is positive at Q and RB. Force is negativ at O, X and TC. Forces is maximum (magnitude) at O, R and XD. Equlibrium is indicated at P, S and Z
Answer» Correct Answer - A::B::C::D

Discussion

898.	A particle is moving along x-axis under the action of a force F, which varies with its position `(x)as fprop(1)/(sqrtx).`Find the variation of power due to this force with x.
Answer» `Fpropx^(-1//2)thereforea=Kx^(-1//2)` `a or (vdv)/(dx)=kx^(-1//2)` ltbr"gt `intvdv=kintx^(-1//2)dx` Find v. `P=Fv` `Pprop(v)/(sqrtx)` `F propx^(1//2)` `a prop^(-1//2)` `a=kx^(-1//2)` (k is a proporationality constant) `(dv)/(dt)=(dx)/(dt)(dv)/(dx)=kx^(-1//2)` `v(dv)/(dx)=kx^(-1//2)` `intvdv=kintx^(-1//2)dx` `(v^(2))/(2)=k(x^(1//2))/(1//2)` `v^(2)propx^(1//2)` `v propx^(1//4)` `P=Fv` `Pprop(x^(1//4))/(sqrt4)` `P propx^(-1//4)`

Discussion

899.	A particle is moving along x-axis under the action of force, F which varies with its position as `F prop(1)/(4sqrtx).` Find the variation of power due to this fore with x.
Answer» `F propx^(-1//4)` `impliesaprop x^(-1//4)` `impliesa=kx^(-1//4)` (where k is a proporationality constant) `a=(dv)/(dt)=(dx)/(dt)(dv)/(dx)=kx^(-1//4)` `(vdv)/(dx)=kx^(-1//4)` `intvdv=kintx^(-1//4)dx` `(v^(2))/(2)=k(x^(3//4))/(3//4).` `v^(2)propx^(3//4)` `v propx^(3//8)` `thereforeP=Fv` `p prop(x^(3//8))/(x^(1//4))P propx^(1//8)`

Discussion

900.	A particle of mass 100 g, is made to describe a vertical circle of radius 1 m. Its instantaneous speed is `1ms^(-1)` when the string makes an sngle of `30^(@)` with the vertical Find the tension in the string at this position. Can the particle complete its circular path? `(g=10ms^(-2))`
Answer» Tension in the string when it make an angle `theta` with the vertical is `T=(mv^(2))/®+mg cos theta` Let speed at the lowest point be `v_(1)` `v_(1)^(2)=v^(2)+2gr(1-costheta)` `(v_(1))_(min)=sqrt(5gr)"chect if"v_(1)lt(v_(1))_(min).` If yes, then it would not complete its vertical circular path. The tension in the string. when it makes and angle `theta` with the vertical is `T=(mv^(2))/(r)+mgcostheta` `=(0.1xx1^(2))/(1)+0.1xx10xx0.866=0.966N` let the speed at the lowest point be `v_(1)` `v_(1)^(2)=v^(2).+2gr(1-costheta)` `=1^(2)+2xx10xx1xx(1-0.866)` `=1+20xx0.134` `=3.68` `v_(1)=sqrt3.68=1.91ms^(-1)` `(v_(1))_(min)=sqrt(5gr)=sqrt(5xx10xx1)=7.07ms^(-1)` `v_(1)lt(v_(1))_(min)implies` the particle would not be able to complete its circular path.

Discussion

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