Question

Two bodies of different masses mi and m2 are
dropped simultaneously from height h, and h2 res-
pectively. The ratio of times taken by these is -
(A) VhVh2 (B) h : họ
(C) h12: h22
(D) m2 h:mh2​

Answer 1

Answer:

Option A is your answer _______________________________

$\huge{ \underline{ \large{ \rm{ \pink{Given:}}}}}$

• Two bodies of different masses m1 and m2
• They dropped from height h1 and h2

$\huge{ \underline{ \rm{ \large{ \red{Find:}}}}}$

• Ratios of time taken??

$\huge{ \underline{ \rm{ \green{ \large{Solution:}}}}}$

$\to{ \sf{M1 : h2 = \frac{1}{2} \times \frac{p1}{m2} ({t1)}^{2} }}$

$\to{ \sf{ M2:h2 = \frac{1}{2} \times \frac{p2}{m2} ({t2})^{2} }}$

Where P1 = m1g, P2 = m2g (weights)

So, let's divide the equations:-

${ \implies{ \sf{( \frac{h1}{h2} ) = \frac{(m1 \times g \times m2)}{(m2 \times g \times m1)} \times {( \frac{t1}{t2} )}^{2} }}}$

${ \implies{ \sf{( \frac{h1}{h2} ) = { (\frac{t1}{t2}) }^{2} }}}$

${ \implies{ \sf{ \frac{t1}{t2} = \sqrt{ (\frac{h1}{h2}) } }}}$

$\: \:$

${ \therefore{ \pink{ \sf{ t_{1}: t_{2} = \sqrt{ h_{1} }: \sqrt{ h_{2}} }}}}$

Answer 2

$\huge\tt{\green{Given : }}$

• Two bodies of different masses m1 and m2
• They dropped from height h1 and h2

$\huge\tt{\red{To \: Find : }}$

• Ratios of time taken.

$\huge\tt {\orange{Solution:}}$

$:\implies{ \sf{M1 : h2 = \dfrac{1}{2} \times \dfrac{p1}{m2} ({t1)}^{2} }}$

$:\implies { \sf{ M2:h2 = \dfrac{1}{2} \times \dfrac{p2}{m2} ({t2})^{2} }}$

On dividing the equations :

${: \implies{ \sf{( \dfrac{h1}{h2} ) = \dfrac{(m1 \times g \times m2)}{(m2 \times g \times m1)} \times {( \dfrac{t1}{t2} )}^{2} }}}$

${ :\implies{ \sf{( \dfrac{h1}{h2} ) = { (\dfrac{t1}{t2}) }^{2} }}}$

${ :\implies{ \sf{ \dfrac{t1}{t2} = \sqrt{ (\dfrac{h1}{h2}) } }}}$

${ \therefore{\underline{\boxed{ \sf{ t_{1}: t_{2} = \sqrt{ h_{1} }: \sqrt{ h_{2}} }}}}}$

________________________________________

Option A is the answer !