*I would never die for my belief because I might be wrong*.

– Bertrand Russell

Two questions on gravitation were included in the IIT-JEE 2010 question paper. They are given below with solution. The first question is a ‘*single correct choice type*’ (single answer type multiple choice) question where as the second question is an ‘*integer type*’ question which has a single digit integer ranging from 0 to 9 as answer.

(1) A thin uniform annular disc (see figure) of mass *M* has outer radius 4*R* and inner radius 3*R*. The work required to take a unit mass from point P on its axis to infinity is

(A) (2*GM* /7*R*) (4√2 – 5)

(B) – (2*GM* /7*R*) (4√2 – 5)

(C) *GM* /4*R*

(D) (2*GM* /5*R*) (√2 – 1)

The work required to take a unit mass from infinity to point P is the gravitational potential at P and it will be negative. Therefore, the work required to take a unit mass from point P to infinity will be numerically equal to the gravitational potential at P but it will be *positive*.

The annular disc can be considered to be made of a large number of concentric rings of radii ranging from 3*R* to 4*R*. Consider one such ring of radius *r* and small thickness *dr*. Its mass *dm* is given by

*dm = *[*M/*π(16*R*^{2 }– 9*R*^{2})] 2π*r dr* = 2*Mrdr*/7*R*^{2}

If the rim of the ring is at distance* x* from the point P, the gravitational potential (*dV*) at P due to the ring is given by

*dV* = – *Gdm/x = *– (2G*M*/7*R*^{2})(*rdr*/*x*)

But *x = *(16R^{2} + r^{2})^{1/2} and so we have *dx= *(½ )×(16R^{2} + r^{2})^{–}^{1/2} ×2*rdr*

Or, *dx = rdr/*(16R^{2} + r^{2})^{1/2} = *rdr/x*

Therefore *dV* = – (2*GM*/7*R*^{2})*dx*

The gravitational potential at P due to the entire annular disc is given by

*V = *∫ *dV = *–* *∫(2*GM*/7*R*^{2})*dx = *– (2G*M*/7*R*^{2})[*x*]

The integration is between the appropriate limits of *x*, which are (16*R*^{2} + 9*R*^{2})^{1/2}* *and (16*R*^{2} + 16*R*^{2})^{1/2},^{ } on substituting the values *r = *3*R* and *r = *4*R* respectively in the expression, *x = *(16R^{2} + r^{2})^{1/2}. Therefore we have^{}

* V *= – (2*GM*/7*R*^{2})[(16*R*^{2} + 16*R*^{2})^{1/2 }– (16*R*^{2} + 9*R*^{2})^{1/2}]

= – (2*GM*/7*R*^{2})[*R*√(32) –* R*√(25)]

Thus *V* = – (2*GM*/7*R*) (4√2 – 5)

The negative sign is to be dispensed with to obtain the work required to take a unit mass from point P to infinity and the correct choice is (2*GM*/7*R*) (4√2 – 5).

(2) Gravitational acceleration on the surface of a planet is (√6/11)*g*, where *g* is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms^{–1}, the escape speed on the surface of the planet in kms^{–1} will be :

[This is is an ‘*integer type*’ question. You have to work out this question and the single digit integer answer is to be shown in the Objective Response Sheet (ORS) by darkening the appropriate bubble].

We have *v*_{esc} = √(2*gR*) where *g* is the acceleration due to gravity and *R* is the radius of the planet (or earth or star).

Therefore* v*_{esc} α √(*gR*).

If *g*_{p }and *g*_{e} are the accelerations due to gravity on the surface of the planet and the earth, *R*_{p} and *R*_{e}* *are their radii, and *g*_{p} and *g*_{e} the escape velocities respectively, we have

*v*_{p}/*v*_{e} = √(* g*_{p}*R*_{p})/* *√(* g*_{e}*R*_{e}) = √[(* g*_{p}/*g*_{e})(*R*_{p}/*R*_{e})] ………(i)

Since the gravitational acceleration on the surface is given by *g =* *GM/R*^{2}* *where *G *is the gravitational constant and *M* is the mass of the planet (or earth or star), we can write

*g =* *G*[(4/3)π*R*^{3}*ρ*]* /R*^{2} where *ρ *is the average mass density of the planet (or earth or star).

Therefore *g* α *Rρ* and *g*_{p}/*g*_{e} = (*R*_{p}/*R*_{e})(*ρ*_{p}/*ρ*_{e}).

From this *R*_{p}/*R*_{e} = (*g*_{p}/*g*_{e})/(*ρ*_{p}/*ρ*_{e}).

But *g*_{p}/*g*_{e} = √6/11 and *ρ*_{p}/*ρ*_{e} = 2/3 as given in the question. Therefore we have

*R*_{p}/*R*_{e} = (√6/11)×3/2

Substituting these in Eq(i),

*v*_{p}/*v*_{e} = √[(√6/11) (√6/11)×3/2] = (√6/11) ×√(3/2) = 3/11

Since *v*_{e} = 11 kms^{–1}, the escape speed on the surface of the planet in kms^{–1} will be

*v*_{p} = 11×(3/11) = 3 kms^{–1}

The answer integer is 3.

i am satisfied wid ur answer

ReplyDeleteIt helped. Thanks :-)

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Deletethank you. i helped me clear the concept 15 years after i have passed my +2 examination. ;)

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