Student Login

Home›Chapters›Gravitation PYQs›Acceleration Due to Gravity›Question 17

MHT CET 2025 (22 Apr, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2025 (22 Apr, Shift 1)

Q17MHT CET 2025 · 22 APR · SHIFT 1Easy

Acceleration Due to Gravity

Time period of a simple pendulum on earth's surface is ' T '. It time period becomes ' xT ' when taken to a height ' $2R$ ' above earth's surface. The value of $x$ will be $(R =$ radius of earth)

Show answer and solution

Answer: (d)

At $h = 2R$ : $g_h = \frac{g}{(1+2)^2} = \frac{g}{9}$ .
$T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = T\sqrt{9} = 3T$
So the answer is (d) $x = 3$ .

Report a mistake

More questions on Acceleration Due to Gravity

A body weighs 48 N on the surface of the earth. The gravitational force experienced by…NEET 2025
The depth at which the value of acceleration due to gravity becomes (1/n) times the value…MHT CET 2025 (25 Apr, Shift 2)
The gravitational pull of the moon is (1/6)th of the earth and mass of moon is (1/8) th…MHT CET 2025 (23 Apr, Shift 2)
The depth at which acceleration due to gravity becomes g/n is ( R = radius of earth, g =…MHT CET 2025 (26 Apr, Shift 2)
Water rises up to height ' x ' in a capillary tube immersed vertically in water. When the…MHT CET 2025 (21 Apr, Shift 1)
The density of a new planet is twice that of earth. The acceleration due to gravity at…MHT CET 2024 (15 May, Shift 2)
The acceleration due to gravity at the surface of the planet is same as that at the…MHT CET 2024 (11 May, Shift 1)
The mass of a planet is 1/10 th that of the earth and its diameter is half that of the…NEET 2024