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 17th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
Two planets A and B are orbiting around the sun. The distances of the two planets A and B from the sun are $r_A$ and $r_B$ respectively. Also $r_B = 225\, r_A$. If the orbital speed of the planet A is 'V' then the orbital speed of planet B will be
A
$\dfrac{V}{3}$
B
$\dfrac{V}{5}$
C
$\dfrac{V}{15}$
D
$\sqrt{15}\,V$
4
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\dfrac{3}{5}$th as much as at present is ($g$ = gravitational acceleration, $R$ = equatorial radius of the earth.)
A
$\sqrt{\dfrac{3}{5}gR}$
B
$\sqrt{\dfrac{2g}{5R}}$
C
$\sqrt{\dfrac{3g}{5R}}$
D
$\sqrt{\dfrac{5R}{2g}}$

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