1
MHT CET 2026 18th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
The lengths of seconds pendulums on the surface of the earth and at an altitude '$h$' from the surface of the earth are $l_s$ and $l_h$ respectively. The radius of the earth is
A
$\dfrac{h\sqrt{l_h}}{\sqrt{l_s}-\sqrt{l_h}}$
B
$\dfrac{h\sqrt{l_h}}{\sqrt{l_h}-\sqrt{l_s}}$
C
$\dfrac{\sqrt{l_h}}{h(\sqrt{l_s}-\sqrt{l_h})}$
D
$\dfrac{\sqrt{l_s}}{h(\sqrt{l_h}-\sqrt{l_s})}$
2
MHT CET 2026 18th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
Time period of a simple pendulum is $T_1$ when on the earth's surface and $T_2$ when taken to a height '2R' above the earth's surface, where 'R' is the radius of the earth. The ratio $T_1 : T_2$ is
A
$1 : 2$
B
$1 : 3$
C
$1 : 4$
D
$1 : 5$
3
MHT CET 2026 18th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth 'h' of the lake is ($\rho$ = density of water in the lake, $g$ = acceleration due to gravity, $P_0$ = atmospheric pressure).
A
$\dfrac{2P_0}{\rho g}$
B
$\dfrac{P_0}{\rho g}$
C
$\dfrac{2P_0}{3\rho g}$
D
$\dfrac{P_0}{2\rho g}$
4
MHT CET 2026 18th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
A spherical object is falling under gravity through a viscous fluid. The sphere attains the terminal velocity when
A
viscous force is zero.
B
buoyant force is equal to force due to gravity
C
viscous force plus force of gravity becomes equal to buoyant force.
D
buoyant force plus viscous force becomes equal to force due to gravity.

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