UPSC Physics Paper 1 - 2025

By Naga Sai RaoJune 22, 20261 min read

Q1.

a. Consider a large stationary cylinder of inner radius R. A smaller solid cylinder of radius r rolls without slipping inside the larger cylinder. Determine the equation of motion of the smaller cylinder.

The key physical idea is that the small cylinder’s center does not move in a straight line: because it rolls on the inside of the fixed cylinder, its center traces a circle of radius $R-r$ . The no-slip condition then ties the orbital motion of the center to the spin of the small cylinder, so the problem is really one of constrained rotation.

Let the small cylinder have mass $m$ , radius $r$ , and moment of inertia about its own center $I$ . Let $\theta$ be the angular position of its center measured from the lowest point of the big cylinder, and let $\phi$ be the spin angle of the small cylinder about its own axis. We assume:

the larger cylinder is fixed,
rolling is without slipping,
gravity acts downward with acceleration $g$ ,
the motion is confined to a vertical plane.

Because the center of the small cylinder moves on a circle of radius $R-r$ , its speed is

v=(R-r)\dot{\theta}.

For rolling without slipping, the point of contact has zero relative velocity, so the arc length traveled by the center along the inner surface equals the arc length unwound by the cylinder’s rotation. With the sign convention chosen so that increasing $\theta$ corresponds to rolling forward, the constraint is

(R-r)\dot{\theta}=r\dot{\phi}.

Now write the kinetic energy. The translational part is

T_{\text{trans}}=\frac12 m (R-r)^2\dot{\theta}^2,

and the rotational part is

T_{\text{rot}}=\frac12 I\dot{\phi}^2 =\frac12 I\left(\frac{R-r}{r}\right)^2\dot{\theta}^2.

So the total kinetic energy is

T=\frac12 (R-r)^2\left(m+\frac{I}{r^2}\right)\dot{\theta}^2.

The height of the center relative to the lowest point is

h=(R-r)(1-\cos\theta),

so the potential energy is

V=mg(R-r)(1-\cos\theta).

Thus the Lagrangian is

L=T-V =\frac12 (R-r)^2\left(m+\frac{I}{r^2}\right)\dot{\theta}^2 -mg(R-r)(1-\cos\theta).

Apply the Euler–Lagrange equation:

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta}=0.

Compute each term:

\frac{\partial L}{\partial \dot{\theta}} =(R-r)^2\left(m+\frac{I}{r^2}\right)\dot{\theta},

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) =(R-r)^2\left(m+\frac{I}{r^2}\right)\ddot{\theta}.

Also,

\frac{\partial L}{\partial \theta} =-mg(R-r)\sin\theta.

Therefore the equation of motion is

(R-r)^2\left(m+\frac{I}{r^2}\right)\ddot{\theta} +mg(R-r)\sin\theta=0.

Dividing through by $(R-r)^2\left(m+\frac{I}{r^2}\right)$ gives

\ddot{\theta} +\frac{g}{R-r}\,\frac{m}{m+I/r^2}\sin\theta=0.

For a solid cylinder,

I=\frac12 mr^2,

m+\frac{I}{r^2}=m+\frac12 m=\frac32 m.

Hence the motion becomes

\ddot{\theta}+\frac{2g}{3(R-r)}\sin\theta=0.

This is the same form as a simple pendulum equation, but with an effective length $\,\frac{3}{2}(R-r)\,$ in the small-angle limit.

Answer: $\boxed{\ddot{\theta}+\frac{2g}{3(R-r)}\sin\theta=0}$ for a solid cylinder rolling without slipping inside a fixed cylinder of inner radius $R$ . For small oscillations, this reduces to

\boxed{\ddot{\theta}+\frac{2g}{3(R-r)}\,\theta=0,}

so the small-oscillation angular frequency is

\boxed{\omega=\sqrt{\frac{2g}{3(R-r)}}\,.}

Takeaway: This problem shows how rolling constraints convert a rigid body’s motion into an effective pendulum-like oscillation, with the inertia changing the frequency compared with a point mass. Similar ideas are used in precision mechanical design and in modern robotics, where rolling-contact models help predict vibration, energy loss, and stability in curved-track motion.

b. Derive the expression for the gravitational self-energy of a uniform solid sphere of mass M and radius R.

The idea is to build the sphere up shell by shell and add the work needed to bring each new layer in from infinity. Because gravity is attractive, the work done by gravity is negative, so the final self-energy is negative.

We are given a uniform solid sphere with total mass $M$ and radius $R$ . We are asked to derive its gravitational self-energy. Assume Newtonian gravity and a continuous mass distribution with constant density.

For a sphere of radius $r$ inside the body, the mass already assembled is

M(r)=\frac{4}{3}\pi r^3\rho

with uniform density

\rho=\frac{M}{\frac{4}{3}\pi R^3}.

Now imagine adding a thin spherical shell of thickness $dr$ at radius $r$ . Its mass is

dm=4\pi r^2\rho\,dr.

The gravitational potential energy to bring this shell from infinity and place it around the already-built interior mass $M(r)$ is

dU=-\frac{G\,M(r)\,dm}{r}.

Substitute the expressions for $M(r)$ and $dm$ :

dU=-\frac{G}{r}\left(\frac{4}{3}\pi r^3\rho\right)\left(4\pi r^2\rho\,dr\right).

Simplify:

dU=-\frac{16}{3}\pi^2 G\rho^2 r^4\,dr.

Integrate from $r=0$ to $r=R$ :

U=-\frac{16}{3}\pi^2 G\rho^2\int_0^R r^4\,dr

U=-\frac{16}{3}\pi^2 G\rho^2\left(\frac{R^5}{5}\right).

U=-\frac{16}{15}\pi^2 G\rho^2 R^5.

Now write $\rho$ in terms of $M$ and $R$ :

\rho=\frac{3M}{4\pi R^3}.

Then

U=-\frac{16}{15}\pi^2 G\left(\frac{3M}{4\pi R^3}\right)^2R^5.

Simplify:

U=-\frac{16}{15}\pi^2 G\cdot \frac{9M^2}{16\pi^2R^6}\cdot R^5 =-\frac{3}{5}\frac{GM^2}{R}.

Answer: $\boxed{U=-\frac{3GM^2}{5R}}$

This result shows that a uniform sphere is more tightly bound the larger its mass and the smaller its radius, with the characteristic scaling $U\propto -M^2/R$ . In real astrophysics, the same idea helps estimate binding energies of stars, planets, and protostellar clouds, and it underlies modern modeling of how gravity competes with pressure in collapse, formation, and compact-object structure.

Takeaway: The self-energy is the amount of gravitational binding that must be overcome to disperse the sphere to infinity, so it is directly connected to stability and collapse. Similar energy accounting is used today in simulations of star formation, planetary differentiation, and gravitational binding in dense astrophysical objects where Newtonian intuition is still a useful first estimate.

c. A particle of rest mass 1 kg and velocity of magnitude 0.9c collides with a particle of mass 2 kg at rest. After collision the two particles coalesce and form a single particle of mass M and velocity V. Determine M and V.

We are given a particle of rest mass $m_1 = 1\,\text{kg}$ moving with speed $u = 0.9c$ , and a second particle of rest mass $m_2 = 2\,\text{kg}$ initially at rest. They collide and stick together, forming one composite particle of rest mass $M$ moving with speed $V$ . We want $M$ and $V$ .

The key physical idea is that in special relativity we must conserve the full energy-momentum 4-vector, not just classical momentum. When two bodies coalesce, some kinetic energy becomes internal energy, so the final rest mass $M$ is larger than $m_1+m_2$ .

For each particle, the relativistic factors are

\gamma = \frac{1}{\sqrt{1-u^2/c^2}}.

For the first particle, with $u=0.9c$ ,

\gamma_1 = \frac{1}{\sqrt{1-0.9^2}}=\frac{1}{\sqrt{0.19}} \approx 2.294.

So its energy and momentum are

E_1 = \gamma_1 m_1 c^2, \qquad p_1 = \gamma_1 m_1 u.

Using $m_1=1\,\text{kg}$ and $u=0.9c$ ,

E_1 \approx 2.294\,c^2, \qquad p_1 \approx 2.294(0.9)c = 2.065\,c

in units consistent with kg and $c$ . The second particle is at rest, so

E_2 = m_2 c^2 = 2c^2, \qquad p_2=0.

Thus the total initial energy and momentum are

E_{\text{tot}} = (\gamma_1 m_1 + m_2)c^2, \qquad p_{\text{tot}} = \gamma_1 m_1 u.

After the collision, the composite particle has rest mass $M$ and speed $V$ , so

E_{\text{tot}} = \Gamma M c^2, \qquad p_{\text{tot}} = \Gamma M V,

where

\Gamma = \frac{1}{\sqrt{1-V^2/c^2}}.

A very efficient way to find the invariant rest mass $M$ is to use

M^2 c^4 = E_{\text{tot}}^2 - p_{\text{tot}}^2 c^2.

Substitute the initial totals:

M^2 c^4 = \left[(\gamma_1 m_1 + m_2)c^2\right]^2 - \left[\gamma_1 m_1 u\right]^2 c^2.

Since $u=0.9c$ , this becomes

M^2 = (\gamma_1 m_1 + m_2)^2 - (\gamma_1 m_1)^2 \frac{u^2}{c^2}.

With $m_1=1$ , $m_2=2$ , $\gamma_1 \approx 2.294$ , and $u^2/c^2=0.81$ ,

M^2 = (2.294+2)^2 - (2.294)^2(0.81).

Compute each term:

(4.294)^2 \approx 18.44, \qquad (2.294)^2(0.81) \approx 4.26.

M^2 \approx 14.18, \qquad M \approx 3.77\,\text{kg}.

Now find $V$ from momentum conservation:

p_{\text{tot}} = \Gamma M V.

A cleaner relation is

V = \frac{p_{\text{tot}}c^2}{E_{\text{tot}}}.

Thus

V = \frac{\gamma_1 m_1 u}{\gamma_1 m_1 + m_2}.

Substitute the values:

V = \frac{(2.294)(1)(0.9c)}{2.294+2}.

V \approx \frac{2.065c}{4.294} \approx 0.481c.

Answer: $\mathbf{M \approx 3.77\ \text{kg}}$ , $\mathbf{V \approx 0.481\,c}$ .

Takeaway: In a relativistic inelastic collision, the final rest mass is not just the sum of the original masses; it includes the kinetic energy converted into internal energy. This is exactly the kind of mass-energy bookkeeping used in high-energy particle collisions and in modern detector analyses, where the invariant mass of a combined system is a primary observable.

d. In a double slit Fraunhofer diffraction experiment, the slit width is 0.12 mm and the spacing between the two slits is 0·48 mm. The distance of the screen from the slits is 1·5 m. If the wavelength of the light used is 600 nm, determine (i) the missing orders of the interference maxima, and (ii) the distance between the central maxima and the first minima.

We are given a double-slit Fraunhofer diffraction setup with

Slit width: $a = 0.12\ \text{mm}$
Slit separation: $d = 0.48\ \text{mm}$
Screen distance: $D = 1.5\ \text{m}$
Wavelength: $\lambda = 600\ \text{nm}$

We are asked to find:

the missing orders of the interference maxima,
the distance from the central maximum to the first diffraction minimum.

The key idea is that in a double-slit pattern, the bright interference fringes are modulated by a single-slit diffraction envelope. A bright interference fringe disappears whenever it falls exactly at a diffraction minimum.

For interference maxima, the condition is

d\sin\theta = m\lambda

where $m = 0,1,2,\dots$ is the order number.

For diffraction minima of each slit, the condition is

a\sin\theta = n\lambda

where $n = 1,2,3,\dots$ .

A missing order occurs when both happen at the same angle, so

d\sin\theta = m\lambda,\qquad a\sin\theta = n\lambda.

Dividing the two equations gives

\frac{d}{a} = \frac{m}{n}.

Now substitute the given slit dimensions:

\frac{d}{a} = \frac{0.48}{0.12} = 4.

\frac{m}{n} = 4 \quad \Rightarrow \quad m = 4n.

That means the interference maxima whose orders are multiples of 4 coincide with diffraction minima and therefore vanish.

So the missing orders are

m = 4,\,8,\,12,\,\dots

Next, the first diffraction minimum is given by $n=1$ :

a\sin\theta = \lambda.

For small angles, $\sin\theta \approx \tan\theta \approx y/D$ , so the position on the screen is

y = D\sin\theta \approx D\frac{\lambda}{a}.

Substitute the values:

y = 1.5 \times \frac{600\times 10^{-9}}{0.12\times 10^{-3}}.

Compute the ratio:

\frac{600\times 10^{-9}}{0.12\times 10^{-3}} = \frac{600}{0.12}\times 10^{-6+3} = 5000\times 10^{-3} = 5\times 10^{-3}.

Thus,

y = 1.5 \times 5\times 10^{-3} = 7.5\times 10^{-3}\ \text{m}.

So the distance from the central maximum to the first minimum is

y = 7.5\ \text{mm}.

Answer: (i) The missing orders are 4th, 8th, 12th, ..., i.e. every multiple of 4. (ii) The distance between the central maximum and the first minimum is $7.5\ \text{mm}$ .

Takeaway: The missing orders appear because the interference maxima can land exactly where the single-slit diffraction pattern has zero intensity, which is a classic example of how wave superposition and diffraction combine. This same principle is used in precision optical metrology and in modern photonics systems where fringe visibility and beam shaping matter, such as interferometric sensors and diffraction-based optical design.

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Naga Sai Rao

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Physics

UPSC Physics Paper 1 - 2025

By Naga Sai RaoJune 22, 20261 min read

Q1.

the larger cylinder is fixed,
rolling is without slipping,
gravity acts downward with acceleration $g$ ,
the motion is confined to a vertical plane.

Because the center of the small cylinder moves on a circle of radius $R-r$ , its speed is

v=(R-r)\dot{\theta}.

(R-r)\dot{\theta}=r\dot{\phi}.

Now write the kinetic energy. The translational part is

T_{\text{trans}}=\frac12 m (R-r)^2\dot{\theta}^2,

and the rotational part is

T_{\text{rot}}=\frac12 I\dot{\phi}^2 =\frac12 I\left(\frac{R-r}{r}\right)^2\dot{\theta}^2.

So the total kinetic energy is

T=\frac12 (R-r)^2\left(m+\frac{I}{r^2}\right)\dot{\theta}^2.

The height of the center relative to the lowest point is

h=(R-r)(1-\cos\theta),

so the potential energy is

V=mg(R-r)(1-\cos\theta).

Thus the Lagrangian is

L=T-V =\frac12 (R-r)^2\left(m+\frac{I}{r^2}\right)\dot{\theta}^2 -mg(R-r)(1-\cos\theta).

Apply the Euler–Lagrange equation:

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta}=0.

Compute each term:

\frac{\partial L}{\partial \dot{\theta}} =(R-r)^2\left(m+\frac{I}{r^2}\right)\dot{\theta},

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) =(R-r)^2\left(m+\frac{I}{r^2}\right)\ddot{\theta}.

Also,

\frac{\partial L}{\partial \theta} =-mg(R-r)\sin\theta.

Therefore the equation of motion is

(R-r)^2\left(m+\frac{I}{r^2}\right)\ddot{\theta} +mg(R-r)\sin\theta=0.

Dividing through by $(R-r)^2\left(m+\frac{I}{r^2}\right)$ gives

\ddot{\theta} +\frac{g}{R-r}\,\frac{m}{m+I/r^2}\sin\theta=0.

For a solid cylinder,

I=\frac12 mr^2,

m+\frac{I}{r^2}=m+\frac12 m=\frac32 m.

Hence the motion becomes

\ddot{\theta}+\frac{2g}{3(R-r)}\sin\theta=0.

This is the same form as a simple pendulum equation, but with an effective length $\,\frac{3}{2}(R-r)\,$ in the small-angle limit.

\boxed{\ddot{\theta}+\frac{2g}{3(R-r)}\,\theta=0,}

so the small-oscillation angular frequency is

\boxed{\omega=\sqrt{\frac{2g}{3(R-r)}}\,.}

b. Derive the expression for the gravitational self-energy of a uniform solid sphere of mass M and radius R.

For a sphere of radius $r$ inside the body, the mass already assembled is

M(r)=\frac{4}{3}\pi r^3\rho

with uniform density

\rho=\frac{M}{\frac{4}{3}\pi R^3}.

Now imagine adding a thin spherical shell of thickness $dr$ at radius $r$ . Its mass is

dm=4\pi r^2\rho\,dr.

The gravitational potential energy to bring this shell from infinity and place it around the already-built interior mass $M(r)$ is

dU=-\frac{G\,M(r)\,dm}{r}.

Substitute the expressions for $M(r)$ and $dm$ :

dU=-\frac{G}{r}\left(\frac{4}{3}\pi r^3\rho\right)\left(4\pi r^2\rho\,dr\right).

Simplify:

dU=-\frac{16}{3}\pi^2 G\rho^2 r^4\,dr.

Integrate from $r=0$ to $r=R$ :

U=-\frac{16}{3}\pi^2 G\rho^2\int_0^R r^4\,dr

U=-\frac{16}{3}\pi^2 G\rho^2\left(\frac{R^5}{5}\right).

U=-\frac{16}{15}\pi^2 G\rho^2 R^5.

Now write $\rho$ in terms of $M$ and $R$ :

\rho=\frac{3M}{4\pi R^3}.

Then

U=-\frac{16}{15}\pi^2 G\left(\frac{3M}{4\pi R^3}\right)^2R^5.

Simplify:

U=-\frac{16}{15}\pi^2 G\cdot \frac{9M^2}{16\pi^2R^6}\cdot R^5 =-\frac{3}{5}\frac{GM^2}{R}.

Answer: $\boxed{U=-\frac{3GM^2}{5R}}$

For each particle, the relativistic factors are

\gamma = \frac{1}{\sqrt{1-u^2/c^2}}.

For the first particle, with $u=0.9c$ ,

\gamma_1 = \frac{1}{\sqrt{1-0.9^2}}=\frac{1}{\sqrt{0.19}} \approx 2.294.

So its energy and momentum are

E_1 = \gamma_1 m_1 c^2, \qquad p_1 = \gamma_1 m_1 u.

Using $m_1=1\,\text{kg}$ and $u=0.9c$ ,

E_1 \approx 2.294\,c^2, \qquad p_1 \approx 2.294(0.9)c = 2.065\,c

in units consistent with kg and $c$ . The second particle is at rest, so

E_2 = m_2 c^2 = 2c^2, \qquad p_2=0.

Thus the total initial energy and momentum are

E_{\text{tot}} = (\gamma_1 m_1 + m_2)c^2, \qquad p_{\text{tot}} = \gamma_1 m_1 u.

After the collision, the composite particle has rest mass $M$ and speed $V$ , so

E_{\text{tot}} = \Gamma M c^2, \qquad p_{\text{tot}} = \Gamma M V,

where

\Gamma = \frac{1}{\sqrt{1-V^2/c^2}}.

A very efficient way to find the invariant rest mass $M$ is to use

M^2 c^4 = E_{\text{tot}}^2 - p_{\text{tot}}^2 c^2.

Substitute the initial totals:

M^2 c^4 = \left[(\gamma_1 m_1 + m_2)c^2\right]^2 - \left[\gamma_1 m_1 u\right]^2 c^2.

Since $u=0.9c$ , this becomes

M^2 = (\gamma_1 m_1 + m_2)^2 - (\gamma_1 m_1)^2 \frac{u^2}{c^2}.

With $m_1=1$ , $m_2=2$ , $\gamma_1 \approx 2.294$ , and $u^2/c^2=0.81$ ,

M^2 = (2.294+2)^2 - (2.294)^2(0.81).

Compute each term:

(4.294)^2 \approx 18.44, \qquad (2.294)^2(0.81) \approx 4.26.

M^2 \approx 14.18, \qquad M \approx 3.77\,\text{kg}.

Now find $V$ from momentum conservation:

p_{\text{tot}} = \Gamma M V.

A cleaner relation is

V = \frac{p_{\text{tot}}c^2}{E_{\text{tot}}}.

Thus

V = \frac{\gamma_1 m_1 u}{\gamma_1 m_1 + m_2}.

Substitute the values:

V = \frac{(2.294)(1)(0.9c)}{2.294+2}.

V \approx \frac{2.065c}{4.294} \approx 0.481c.

Answer: $\mathbf{M \approx 3.77\ \text{kg}}$ , $\mathbf{V \approx 0.481\,c}$ .

We are given a double-slit Fraunhofer diffraction setup with

Slit width: $a = 0.12\ \text{mm}$
Slit separation: $d = 0.48\ \text{mm}$
Screen distance: $D = 1.5\ \text{m}$
Wavelength: $\lambda = 600\ \text{nm}$

We are asked to find:

the missing orders of the interference maxima,
the distance from the central maximum to the first diffraction minimum.

For interference maxima, the condition is

d\sin\theta = m\lambda

where $m = 0,1,2,\dots$ is the order number.

For diffraction minima of each slit, the condition is

a\sin\theta = n\lambda

where $n = 1,2,3,\dots$ .

A missing order occurs when both happen at the same angle, so

d\sin\theta = m\lambda,\qquad a\sin\theta = n\lambda.

Dividing the two equations gives

\frac{d}{a} = \frac{m}{n}.

Now substitute the given slit dimensions:

\frac{d}{a} = \frac{0.48}{0.12} = 4.

\frac{m}{n} = 4 \quad \Rightarrow \quad m = 4n.

That means the interference maxima whose orders are multiples of 4 coincide with diffraction minima and therefore vanish.

So the missing orders are

m = 4,\,8,\,12,\,\dots

Next, the first diffraction minimum is given by $n=1$ :

a\sin\theta = \lambda.

For small angles, $\sin\theta \approx \tan\theta \approx y/D$ , so the position on the screen is

y = D\sin\theta \approx D\frac{\lambda}{a}.

Substitute the values:

y = 1.5 \times \frac{600\times 10^{-9}}{0.12\times 10^{-3}}.

Compute the ratio:

\frac{600\times 10^{-9}}{0.12\times 10^{-3}} = \frac{600}{0.12}\times 10^{-6+3} = 5000\times 10^{-3} = 5\times 10^{-3}.

Thus,

y = 1.5 \times 5\times 10^{-3} = 7.5\times 10^{-3}\ \text{m}.

So the distance from the central maximum to the first minimum is

y = 7.5\ \text{mm}.

Answer: (i) The missing orders are 4th, 8th, 12th, ..., i.e. every multiple of 4. (ii) The distance between the central maximum and the first minimum is $7.5\ \text{mm}$ .

upsc-physics-paper-1

Naga Sai Rao

Some things fade fast. Some last. Learn the ones that last.

About the author →