1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 30 InterviewSolution

1451.	A ball bounce of `80%` of its original height . What fraction of its mechanical energy is lost in each bounce ?
Answer» Suppose the ball is dropped from a height `h`. Initial `K.E.=mgh`. P.E. after the bounce `=mgxx(80%)h=(4)/(5)mgh` P.E. lost in each bounce `=mgh -(4)/(5)mgh=(1)/(5)mgh` Fraction of P.E. lost in each bounce `=((1)/(5)mgh)/(mgh)=0.20`

Discussion

1452.	The energy possessed by the body by virtue of its position is called as(a) potential energy (b) kinetic energy (c) mechanical energy (d) none
Answer» (a) potential energy

Discussion

1453.	The energy possessed by a body due to its motion is called as (a) potential energy (b) kinetic energy (c) mechanical energy (d) none
Answer» (b) kinetic energy

Discussion

1454.	A truck of mass 1000 kg accelerates uniformly from rest to a velocity of 15 ms-1 in 5 seconds. Calculate (i) its acceleration, (ii) its gain in K.E., (iii) average power of the engine during this period, neglect friction.
Answer» (i) \(a = \frac{v-u}{t}\) = 3m/s² (ii) Gain in K.E. = \(\frac{1}{2}\)m(v²- u²) = 1.125 x 10⁵ J (iii) P = \(\frac{W}{t}\) = 22500W.

1454.

A truck of mass 1000 kg accelerates uniformly from rest to a velocity of 15 ms-1 in 5 seconds. Calculate (i) its acceleration, (ii) its gain in K.E., (iii) average power of the engine during this period, neglect friction.

Answer»

(i) \(a = \frac{v-u}{t}\) = 3m/s²

(ii) Gain in K.E. = \(\frac{1}{2}\)m(v²- u²) = 1.125 x 10⁵ J

(iii) P = \(\frac{W}{t}\) = 22500W.

Discussion

1455.	A ball at rest is dropped from a height of 12 m. It loses 25% of its kinetic energy in striking the ground, find the height to which it bounces. How do you account for the loss in kinetic energy?
Answer» If ball bounces to height h’ , then mgh’ = 75% of mgh ∴ h’ = 0.75 h = 9 m.

1455.

A ball at rest is dropped from a height of 12 m. It loses 25% of its kinetic energy in striking the ground, find the height to which it bounces. How do you account for the loss in kinetic energy?

Answer»

If ball bounces to height h’ , then

mgh’ = 75% of mgh

∴ h’ = 0.75 h = 9 m.

Discussion

1456.	A ball bounces to 80% of its original height. Calculate the mechanical energy lost in each bounce.
Answer» Let Initial P.E. = mgh P.E. after first bounce = mg × 80% of h = 0.80 mgh P.E. lost in each bounce = 0.20 mgh ∴ Fraction of P.E. lost in each bounce = \(\frac{0.20 mgh}{mgh}\) = 0.20

1456.

A ball bounces to 80% of its original height. Calculate the mechanical energy lost in each bounce.

Answer»

Let Initial P.E. = mgh

P.E. after first bounce = mg × 80% of h = 0.80 mgh

P.E. lost in each bounce = 0.20 mgh

∴ Fraction of P.E. lost in each bounce = \(\frac{0.20 mgh}{mgh}\) = 0.20

Discussion

1457.	The work done by the goal keeper catches the ball coming towards him by applying a force is (a) positive (b) negative (c) zero(d) infinity
Answer» (b) negative

Discussion

1458.	A shell of mass `200g` is ejected from a gun of mass `4 kg` by an explosion that generate `1.05 kJ` of energy. The initial velocity of the shell isA. `100ms^(-1)`B. `80ms^(-1)`C. `40ms^(-1)`D. `120ms^(-1)`
Answer» Correct Answer - a

Discussion

1459.	A ball of mass 1 kg and another of mass 2 kg are dropped from a tall building whose height is 80 m. After, a fall of 40 m each towards Earth, their respective kinetic energies will be in the ratio of(a) \(\sqrt{2}\) : 1 (b) 1 :\(\sqrt{2}\) (c) 2 : 1 (d) 1 : 23
Answer» Correct answer is (d) 1 : 23

Discussion

1460.	An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water of the jet. What is the rate at which kinetic energy is imparted to water?(a) \(\frac{1}{2}\) mv2(b) mv3(c) \(\frac{1}{3}\) mv2(d) \(\frac{1}{4}\) mv2
Answer» Correct answer is (a) \(\frac{1}{2}\)mv²

Discussion

1461.	An explosion blows a rock into three parts. Two parts go off at right angles to each other . These two are `1 kg` first part moving with a velocity of `12 ms^(-1) and 2 kg` second part moving with a velocity of `8 ms^(-1)`. If the third part flies off with a velocity of `4 ms^(-1)`. Its mass would beA. 5KgB. 7KgC. 17KgD. 3Kg
Answer» Correct Answer - a

Discussion

1462.	A body of mass 1 kg is thrown upwards with a velocity 20 ms-1 . It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?(a) 20 J (b) 30 J (c) 40 J(d) 10 J
Answer» Correct answer is (a) 20 J

Discussion

1463.	A body of mass `1 kg` is thrown upwards with a velocity `20 ms^(-1)`. It momentarily comes to rest after attaining a height of `18 m`. How much energy is lost due to air friction? `(g = 10 ms^(-2))`A. 20JB. 30JC. 40JD. 10J
Answer» Correct Answer - a

Discussion

1464.	A mass of 2.5 kg is hung on a steel wire. This increases 0.25 cm in its length. Determine the work done to pull the wire. (g = 10 m/s2) [Hint: Mg = kx, W = \(\frac{1}{2}\) kx2]
Answer» Given; F = 2.5 xg = 2.5 × 10 = 25 N x = 0.25 cm = 0.25 x 10^-2 m = 2.5 x 10^-3 m If the force constant of the wire be k, then \(F=k x \Rightarrow k=\frac{F}{x}=\frac{25}{2.5 \times 10^{-3}}=10^{4} \mathrm{N} \cdot \mathrm{m}^{-1}\) ∴ The work done in pulling the wire, W = \(\frac{1}{2}\) kx²= \(\frac{1}{2}\) × 10⁴ × 2.5 x 2.5 x 10^-6 = \(\frac{1}{2}\)× 6.25 × 10^-2 = 3.125 x 10^-2 J = 0.03125 J

1464.

A mass of 2.5 kg is hung on a steel wire. This increases 0.25 cm in its length. Determine the work done to pull the wire. (g = 10 m/s2) [Hint: Mg = kx, W = \(\frac{1}{2}\) kx2]

Answer»

Given; F = 2.5 xg = 2.5 × 10 = 25 N
x = 0.25 cm = 0.25 x 10^-2 m = 2.5 x 10^-3 m
If the force constant of the wire be k, then
\(F=k x \Rightarrow k=\frac{F}{x}=\frac{25}{2.5 \times 10^{-3}}=10^{4} \mathrm{N} \cdot \mathrm{m}^{-1}\)

∴ The work done in pulling the wire,
W = \(\frac{1}{2}\) kx²= \(\frac{1}{2}\) × 10⁴ × 2.5 x 2.5 x 10^-6
= \(\frac{1}{2}\)× 6.25 × 10^-2
= 3.125 x 10^-2 J = 0.03125 J

Discussion

1465.	Define kilowatt hour and convert it into joules.
Answer» Kilowatt hour : “When an electric power of one kilowatt flows through a conductor for one hour, then electrical energy which flows through the conductor is one kilowatt hour.” 1KWH = 1000W x 1h 1KWH = 1000W x (60 x 60)sec = 1000J/S x 3600s = 100000J 1KWH = 10⁵J

1465.

Define kilowatt hour and convert it into joules.

Answer»

Kilowatt hour : “When an electric power of one kilowatt flows through a conductor for one hour, then electrical energy which flows through the conductor is one kilowatt hour.”

1KWH = 1000W x 1h

1KWH = 1000W x (60 x 60)sec

= 1000J/S x 3600s = 100000J

1KWH = 10⁵J

Discussion

1466.	A small block of mass `1` kg is a circular are of ratius `40` m . The block sides along the track without topping and a frictionnal force acts on it in the direction opposite in the instrmens velocity . The work done in evercoming the friction up to the point `Q` as shown is the figure below is `150 J` (Take the acceleration due to gravity `g = 10 ms^(-2))` The magnitude of the normal reaction that acts on the block at the point `Q` isA. (a) `7.5N`B. (b) `8.6N`C. (c) `11.5N`D. (d) `22.5N`
Answer» Correct Answer - A `N-mg cos 60^@=(mv^2)/(R)` `N=5+5/2=7.5N c`

Discussion

1467.	Force acting on a particle moving in the x-y plane is `vecF=(y^2hati+xhatj)N`, x and y are in metre. As shown in figure, the particle moves from the origin O to point A `(6m, 6m)`. The figure shows three paths, OLA, OMA, and OA for the motion of the particle from O to A. Work done for motion along path OA is nearly A. (a) `383J`B. (b) `90J`C. (c) `180J`D. (d) None of these
Answer» Correct Answer - B `W_(OA)=int F_xdx+intF_ydy` `=underset0overset6inty^2dx+underset0overset6intxdy`, for this path: `y=x` `=underset0overset6intx^2dx+underset0overset6intydy=[x^3/3]_0^6+[y^2/2]_0^6=90J`

Discussion

1468.	Force acting on a particle moving in the x-y plane is `vecF=(y^2hati+xhatj)N`, x and y are in metre. As shown in figure, the particle moves from the origin O to point A `(6m, 6m)`. The figure shows three paths, OLA, OMA, and OA for the motion of the particle from O to A. Which of the following is correct?A. (a) OAB. (b) OMAC. (c) OLAD. (d) Work done has the same value for all the three paths.
Answer» Correct Answer - B (b)

Discussion

1469.	Force acting on a particle moving in the x-y plane is `vecF=(y^2hati+xhatj)N`, x and y are in metre. As shown in figure, the particle moves from the origin O to point A `(6m, 6m)`. The figure shows three paths, OLA, OMA, and OA for the motion of the particle from O to A. Which of the following is correct?A. (a) There is equal probability for the force being conservative or non-conservative.B. (b) Conservative or non-conservative nature of force cannot be predicted on the basis of given information.C. (c) The given force is non-conservative.D. (d) The given force is conservative.
Answer» Correct Answer - C For path `OMA`: Path `OM: vecF=y^2hati, dx=0` `W_(OM)=underset0overset6intvecF.dvecx=0` Path `MA: vecF=(36hati+xhatj)N` `W_(MA)=underset(x_1)overset(x_2)int F_xdx+underset(y_1)overset(y_2)intF_ydy, dy=0` `=36xx6+0=216J` `W_(OMA)=216J` For path OLA: Path `OL: vecF=hatj`, `dy=0` `W_(OL)=0` Path `LA: vecF=(y^2hati+6hatj)N` `W_(LA)=underset(x_1)overset(x_2)int F_xdx+underset(y_1)overset(y_2)int F_ydy` `=0+6xx(y_2-y_1)=6xx6=36J` `:. W_(OMA)!=W_(OLA)` Hence, force is not conservative.

Discussion

1470.	A man is running down an inclined plane. Is he doing work or is work being done on him? Why is it that under normal circumstances when we are running down an inclined plane we feel that we are doing work?
Answer» Work is being done by gravity on the man. The man loses height as he races down the inclined plane. Normally, we feel that we are doing work, as we have to prevent ourselves from moving too fast. In this process, we do work by reducing our kinetic energy resulting from the decrease in height.

Discussion

1471.	Force acting on a particle moving in the x-y plane is `vecF=(y^2hati+xhatj)N`, x and y are in metre. As shown in figure, the particle moves from the origin O to point A `(6m, 6m)`. The figure shows three paths, OLA, OMA, and OA for the motion of the particle from O to A. Now consider another situation. A force `vecF=(4hati+3hatj)N` acts on a particle of mass `2kg`. The particle under the action of this force moves from the origin to a point A `(4m, -8m)`. Initial speed of the particle, i.e., its speed at the origin is `2sqrt6ms^-1`. Figure shows three paths for the motion of the particle from O to A. Speed of the particle at A will be nearlyA. (a) `4.0ms^-1`B. (b) `2.8ms^-1`C. (c) `3.6ms^-1`D. (d) `5.6ms^-1`
Answer» Correct Answer - A `a=((4hati+3hatj))/(2)` `v^2=u^2+2ax` `=4xx6+2((4hati+3hatj))/(2)*(4hati-8hatj)=2A-8=16` `v=4ms^-1`

Discussion

1472.	Force acting on a particle moving in the x-y plane is `vecF=(y^2hati+xhatj)N`, x and y are in metre. As shown in figure, the particle moves from the origin O to point A `(6m, 6m)`. The figure shows three paths, OLA, OMA, and OA for the motion of the particle from O to A. Which of the following is correct?A. (a) There is equal probability for the force being conservative or non-conservative.B. (b) Conservative or non-conservative nature of the force cannot be predicted on the basis of the given information. z.C. (c) The force is non-conservativeD. (d) The force is conservative.
Answer» Correct Answer - D Here `W_(OMA)=W_(OLA)` Hence, force is conservative.

Discussion

1473.	Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative: (a) a chicken scratching the ground, (b) a person studying, (c) a crane lifting a bucket of concrete, (d) the gravitational force on the bucket in part (c), and (e) the leg muscles of a person in the act of sitting down.
Answer» a. Positive work is done by the chicken on the dirt. b. The person does no work on anything in the environment. Perhaps some extra chemical energy goes thorugh energy transmitted electrically and is converted into internal energy in his brain, but it would be very hard to quantify "extra". c. Positive work is done by crane on bucket in lifting it from ground. d. Gravitational force will do negative work because the force is opposite to displacement. e. When we do any activity our forces in muscles will do some positive work.

Discussion

1474.	The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as `K=as^2`. The force acting on the particle isA. (a) `2a(s^2)/(R)`B. (b) `2as[1+(s^2)/(R)]^(1//2)`C. (c) `2as`D. (d) `2a`
Answer» Correct Answer - B Given that `K=as^2` or `1/2mv^2=as^2` or `mv^2=2as^2` (i) Differenciating w.r.t. time, we get `m_2vxx(dv)/(dt)=2axx2sxx(ds)/(dt)` But `(ds)/(dt)=v` So `2m(dv)/(dt)=4as` or `m(dv)/(dt)=2as` Now, `m(dv)/(dt)`=tangential force `=F_t` `F_t=2as` Centripetal force `=F_r(mv^2)/(R)=(2as^2)/(R)` `F_(n et)=sqrt(F_t^2+F_r^2)=sqrt((2as)^2+((2as^2)/(R))^2)` `=2assqrt(1+s^2/R^2)`

Discussion

1475.	Can kinetic energy be negative? Explain.
Answer» No, kinetic energy is always positive. Mass and square of speed are both positive.

Discussion

1476.	A force given by the relation `F=8t`, acts on a body of mass `2 kg`, initialiy at rest. Find the work done by this force on the body during first `2` seconds of its motion.A. `64J`B. zeroC. `-64J`D. none of these
Answer» `F=3t` `m.(dv)/(dt)=8t` `m int_(0)^(v)dv=8 int_(0)^(t) dt` `mv=4t^(2)` or `m(dx)/(dt)=4t^(2)rArrdx=(4t^(2))/(m)dt=2t^(2)dt` `W=intFdx=int_(0)^(2)(8txx2t^(2))dt=64J`

Discussion

1477.	The SI unit of power is watt. It is expressed in terms of mass, length and time as: (a) kgm2s-3 (b) kgms-3 (c) kg2m2s-2 (d) kgms-2
Answer» (a) kg m²s^-3

Discussion

1478.	A spring of force constant `800N//m` has an extension of 5cm. The work done in extending it from 5cm to 15cm isA. 16 JB. 8 JC. 32 JD. 24 J
Answer» Correct Answer - B

Discussion

1479.	The adjoining diagram shows the velocity versus time plot for a particle . The work done by the force on the particel is positive from A. A to BB. B to CC. C to DD. D to E
Answer» Correct Answer - A

Discussion

1480.	A force `vecF=6xhati+2yhatj` displaces a body from `vecr_1=3hati+8hatj` to `vecr_2=5hati-4hatj`. Find the work done by the force.
Answer» Given force `vecF=6xhati+2yhatj` Initial position `vecr_1=3hati+8hatj` and final position `vecr_2=5hati-4hatj` We know work done, `W=underset(s_1)overset(s_2)intvecF.dvecs=underset(s_1)overset(s_2)vecF(dxhati+dyhatj)` `W=underset(s_1)overset(s_2)int(6xhati+2yhatj)(dxhati+dyhatj)` `=underset3overset5int6xdx+underset8overset(-4)int2ydy` `=[3x^2]_3^5+[y^2]_8^-4=[75-27]+[16-64]=0J`

Discussion

1481.	Assertion: A quick collision between two bodies is more violent that show collision , even when initial and final velocity are identical. Reason: The rate of change of momentum determine that force is small or large.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false.
Answer» In a quick collision, time `t` is small. As `F xx t = constant`, therefore, force involved is large, i.e., collision is more violent in comparison to slow collision.

Discussion

1482.	Assertion: A quick collision between two bodies is more violent that show collision , even when initial and final velocity are identical. Reason: The rate of change of momentum determine that force is small or large.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false.
Answer» Correct Answer - A In a quick collision, time`t` is small. As `Fxxt=` constant, therefore, force involved is large,` i.e.,` collision is more violent. Both, the assertion and reason are true and the latter is correct explantion of the former.

Discussion

1483.	A motor drives a body along a straight line with a constant force. The power P developed by the motor muat vary with time t asA. B. C. D.
Answer» Correct Answer - B

Discussion

1484.	A motor drives a body along a straight line with a constant force. The power P developed by the motor muat vary with time t asA. B. C. D.
Answer» Correct Answer - A (a) F=constant `:.` A=constant or v =at Now, `P=F.v=F.at or P prop t` (as F and a both are constant ) Hence,P-t graph is a straight line passing through origin.

Discussion

1485.	Power applied to a particle varices with time as `P =(3t^(2)-2t + 1)` watt, where t is in second. Find the change in its kinetic energy between time `t=2s` and `t = 4 s` .A. 32 JB. 46 JC. 61 JD. 102 J
Answer» Correct Answer - B (b) Change in kinetic energy of the particle `DeltaKE=W=int_(2)^(4)Pdt=int_(2)^(4)(3t^(2)-2t+1)dt=[t^(3)-t^(2)+]_(2)^(4)=46J`

Discussion

1486.	A particle is released from height `H`. At cartain height from the ground its kinetic energy is twice its gravitational potential energy. Find the height and speed of particle at that height.A. `(H)/(3),sqrt((2gH)/(3))`B. `(H)/(3),2sqrt((gH)/(3))`C. `(2H)/(3),sqrt((2gH)/(3))`D. `(H)/(3),sqrt(2gH)`
Answer» Correct Answer - B

Discussion

1487.	What forms of entertainment came up in nineteenth century England to provide leisure activities for the people.
Answer» Forms of entertainment that came up in nineteenth-century England to provide leisure activities for the people were aplenty. For the upper classes, an annual “London Season” was one of the sources of leisure. It comprised the opera, the the are and classical music events. For the working classes, pubs, discussions and meetings for political action served the same purpose. Libraries, art galleries and museums were new types of entertainment brought about through the utilisation of state money. Music halls and cinema theatres too became immensely popular with the lower classes. Industrial workers were encouraged to undertake seaside vacations to rejuvenate from the banes of working in the polluting environment of factories.

Discussion

1488.	If the force F acting on a body as a function of x then the work done in moving a body from x = 1 m to x = 3m is(a) 6 J (b) 4 J (c) 2.5 J(d) 1 J
Answer» Correct answer is (b) 4 J

Discussion

1489.	The work done by the external ajent in stretching a spring of force constant `k` from length `l_(1)`to `l_(2)` isA. `k(l_(2)^(2) - l_(1)^(2))`B. `(1)/(2)k(l_(2)^(2) - l_(1)^(2))`C. `k(l_(2) - l_(1))`D. `k//2(l_(2) - l_(1))`
Answer» Correct Answer - B `U_(2) = (1)/(2)Kl_(2)^(2)` ` U_(1) = (1)/(2)Kl_(2)^(2)` `W=U_(2) - U_(1)` `(1)/(2)Kl_(2)^(2)- (1)/(2)Kl_(1)^(2) = (1)/(2)K(l_(2)^(2)- l_(1)^(2))`

Discussion

1490.	In the figure -3.90 shown, the net work done by the tension when the bigger block of mass M touches the ground is : A. `+Mgd`B. `-(M+m)gd`C. `-Mgd`D. Zero
Answer» Correct Answer - D

Discussion

1491.	A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
Answer» No, work done by a body moving along a closed loop is necessarily zero, only if all the forces acting on the system are conservative.

Discussion

1492.	How high must a body be lifted to gain an amount of P.E. equal to the K.E. it has when moving at speed 20 ms-1. (The value of acceleration due to gravity at a place is 9.8 ms-1).
Answer» mgh = \(\frac{1}{2}\)mv² so h= 20.2 m

1492.

How high must a body be lifted to gain an amount of P.E. equal to the K.E. it has when moving at speed 20 ms-1. (The value of acceleration due to gravity at a place is 9.8 ms-1).

Answer»

mgh = \(\frac{1}{2}\)mv²

so h= 20.2 m

Discussion

1493.	Consider the decay of a free neutron at rest: n`top+e^(-)` Show that the tow-body dacay of this type must necessarily give an electron of fixed energy and, therefore, cannot for the observed continous energy distribution in the `beta`-decay of a neutron or a nucleus.
Answer» Let the masses of the electron and proton be m and M respectively. Let v and V be the velocities of electroni and proton respectively. Using law of conservation of momentum. Momentum of electron + momentum of proton= momentum of neutron `therefore mv+MV=0 rArr V=-m/Mv` Clearly, the electronand the proton move in opposite directions. If mass `Deltam` has been converted into energy in th reaction, then `1/2mv^(2)+1/2MV^(2)= Deltam xx c^(2)` or `1/2mv^(2)+1/2M[-m/M]^(2)v^(2)=Deltamc^(2)` or `1/2mv^(2)[1+m/M]=Deltamc^(2)` or `v^(2)=(2MDeltamc^(2))/(m(M+m))` Thus, it is proved that the value of `v^(2)` is fixed since all the quantities in right hand side are constant it establishes that the emitted electron must have a fixed energy and thus we cannot account for the continous energy distribution in the `beta`-decay of a neutron.

Discussion

1494.	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 ½ (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.

1494.

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 ½ (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

1495.	A partical of mass `m` is driven by a machine that deleveres a constant power `k` watts. If the partical starts from rest the force on the partical at time `t` isA. `sqrt(mK)t^(-1//2)`B. `sqrt(2mK)t^(-1//2)`C. `(1)/(2)sqrt(mK)t^(-1//2)`D. `sqrt((mK)/(2))t^(-1//2)`
Answer» Correct Answer - D Here, power `= K wat t, time =t` as`P=Fupsilon=maupsilon, a=(d upsilon)/(dt)` `K=m (d upsilon)/(dt)` or `upsilond upsilon=(Kdt)/(m)` …(i) Integrating, we get `int upsilon dupsilon=int(Kdt)/(m)` or `(upsilon^(2))/(2)=(K)/(m)intdt` or `(upsilon^(2))/(2)=(K)/(m)t` or `upsilon=sqrt((2Kt)/(m))` Now, `a=(dupsilon)/(dt)=(d)/(dt)(sqrt((2Kt)/(m)))=sqrt((2K)/(m))((1)/(2)t^(-1//2))` `=(1)/(2)sqrt((2K)/(mt))` As `F=ma=mxx(1)/(2)sqrt((2K)/(mt))` or `F=sqrt((mK)/(2t))` or `F=sqrt((mK)/(2))t^(-1//2)`

Discussion

1496.	The block of mass `M` moving on the frictionless horizontal surface collides with the spring constant `k` and compresses it by length `L` . The maximum momention of the block after collision isA. `(kL^(2))/(2 M)`B. `sqrt(Mk) L(c )`SC. `(ML^(2))/(k)`D. zero
Answer» Correct Answer - B (b) `(1)/(2) M nu^(2) = (1)/(2) k L6(2) rArr nu = sqrt(k)/(M) . L` Momentum `= M xx nu = M xx sqrt (k)/(M). L = sqrt(kM) . L`

Discussion

1497.	A block of mass m moving at a speed v compresses a spring throgh a distance x before its speed is halved. Find the spring constant of the spring.
Answer» Correct Answer - B::C::D `E_(i) = E_(f)` `:. (1)/(2)mv^(2) = (1)/(2)m((v)/(2))^(2) + (1)/(2)kx^(2)` or `k = (3v^(2)m)/(4x^(2))`

Discussion

1498.	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): Sign of work done by the camel on the load during parts of motion, accelerated motion, uniform motion and retarded motion, respectively areA. (a) +ve, +ve, +veB. (b) +ve, +ve, -veC. (c) +ve, zero, -veD. (d) +ve, zero, +ve
Answer» Correct Answer - A i. During accelerated motion, negative work is done against friction and there is also change in kinetic energy. Hence, net work needed is positive. ii. During uniform motion, work is done against friction only and that is positive. iii. During restarted motion, the load has to be stopped in exactly `50m`. If only friction is considered, then the load stops in `12.5m` which is less than where it has to stop. Hence, the camel has to apply some force so that the load stops in `50m` `(gt12.5m)`. Therefore, the work done in this case is also positive.

Discussion

1499.	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): Maximum power transmitted by the camel to load isA. (a) `6250Js^-1`B. (b) `5000Js^-1`C. (c) `10^5Js^-1`D. (d) `1250Js^-1`
Answer» Correct Answer - A Maximum force applied by camel is during the accelerated motion. We have `v^2-u^2=-2as` `25=0^2+2xxaxx50` `a=0.25ms^-2` For accelerated motion, `F_C-f=ma` `F_C=mumg+ma` `=0.1xx1000xx10+1000xx0.25` `=1000+250=1250N` This is the critical point just before the point where it attains maximum velocity of almost `5ms^-1`. Hence, maximum power at this point is `1250xx5=6250Js^-1`

Discussion

1500.	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 ratio of magnitude of work done by camel on the load during accelerated motion to retarded motion isA. (a) `3:5`B. (b) `2.2:1`C. (c) `1:1`D. (d) `5:3`
Answer» Correct Answer - D `W_(CL)\|_("accelerated motion")=DeltaKE-W_(f riction)` where `W_(CL)` is work done by camel on the load `=[1/2mv^2-0]-[-mu_kmgxx50]` `=1/2xx1000xx5^2+0.1xx10xx1000xx50` `=1000[125/2]` Similarly, `W_(CL) \|_(ret ardation)=DeltaKE-W_(f riction)` `=[0-1/2mv^2]-[-mu_kmgxx50]` `=1000[75/2]` `(W_(CL)\|_("accelerated motion"))/(W_(CL)\|_("retarded motion"))=125/75=5/3`

Discussion

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