Clock A based on spring oscillations and a clock B based on oscillations of simple pendulum are synchronised on earth. Both are taken to mars whose mass is 0.1 times the mass of earth and radius is half that of earth. Which of the following statement is correct

The correct statement is: Time measured in clock A will be greater than that in clock B.

To understand why this is correct, let's analyze how the time periods of both clocks change when taken from Earth to Mars.

1. Clock A (Spring-Mass System):
The time period of a spring-mass oscillator is given by the formula:
$T=2\pi \sqrt{\frac{m}{k}}$
where:
- $m$ is the mass of the block,
- $k$ is the spring constant.
Notice that this formula does not depend on the acceleration due to gravity ($g$). Therefore, the time period of Clock A remains unchanged when moved to Mars:
$T_{\mathrm{A,\; Mars}}=T_{\mathrm{A,\; Earth}}$

2. Clock B (Simple Pendulum):
The time period of a simple pendulum is given by the formula:
$T=2\pi \sqrt{\frac{L}{g}}$
where:
- $L$ is the length of the pendulum,
- $g$ is the acceleration due to gravity on the planet.
Since the time period is inversely proportional to the square root of $g$ ($T\propto \frac{1}{\sqrt{g}}$), we need to find how $g$ changes on Mars.

3. Gravity on Mars vs. Earth:
The acceleration due to gravity on a planetary body is given by:
$g=\frac{GM}{R^2}$
Given for Mars:
- Mass of Mars, $M_M=0.1M_E$ (where $M_E$ is Earth's mass),
- Radius of Mars, $R_M=0.5R_E$ (where $R_E$ is Earth's radius).
Let's find the ratio of gravity:
$\frac{g_M}{g_E}=\left(\frac{M_M}{M_E}\right)\times {\left(\frac{R_E}{R_M}\right)}^2$
$\frac{g_M}{g_E}=0.1\times {\left(\frac{1}{0.5}\right)}^2=0.1\times 4=0.4$
Since $g_M=0.4g_E$, we have $g_M<g_E$.

4. Comparing Time Measured:
Since gravity on Mars is less than that on Earth ($g_M<g_E$), the time period of Clock B on Mars will increase:
$T_{\mathrm{B,\; Mars}}=2\pi \sqrt{\frac{L}{g_M}}>T_{\mathrm{B,\; Earth}}$
A longer time period means Clock B beats slower (each tick takes more actual time). As a result, Clock B runs slow and records/measures less time for a given interval.
On the other hand, Clock A's time period remains unaffected by gravity ($T_{\mathrm{A,\; Mars}}=T_{\mathrm{A,\; Earth}}$), so it continues to run at its normal speed.
Since Clock B runs slower than its normal rate while Clock A runs at its normal rate, the time measured by Clock A will be greater than the time measured by Clock B over the same physical duration.