Two bodies of different masses $m_{a}$ and $m_{b}$ are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

a:b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b:a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\sqrt{a} : \sqrt{b}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$a^{2} : b^{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$\sqrt{a} : \sqrt{b}$

We have, in general

$h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$

For the two bodies in question, we have

$t_{a} = \sqrt{\frac{2 a}{g}}$ and $t_{b} = \sqrt{\frac{2 b}{g}} \Rightarrow \frac{t_{a}}{t_{b}} = \sqrt{\frac{a}{b}}$

Suggest Corrections

Similar questions

Two bodies of different masses $m_{a}$ and $m_{b}$ are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

Q. Two bodies of different masses

m_{a}

and

m_{b}

are dropped from two different heights

a

and

b

respectively. Ratio of times taken by the two to drop through these distances is:

Two bodies of different masses $m_{a}$ and $m_{b}$ are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

Q. Two bodies of different masses

m_{a}

and

m_{b}

are dropped from two different heights, viz

a

and

b

. The ratio of time taken by the two to drop through these distances is

Q. The bodies of different masses

m_{a}

and

m_{b}

are dropped from two different heights i.e., a and b. The ratio of time taken by both bodies to drop through these distance is :

Join BYJU'S Learning Program

Two bodies of different masses ma and mb are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

The correct option is C √a:√b We have, in general h=12gt2⇒t=√2hg For the two bodies in question, we have ta=√2ag and tb=√2bg⇒tatb=√ab

Two bodies of different masses $m_{a}$ and $m_{b}$ are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

The correct option is C
$\sqrt{a} : \sqrt{b}$

We have, in general

$h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$

For the two bodies in question, we have

$t_{a} = \sqrt{\frac{2 a}{g}}$ and $t_{b} = \sqrt{\frac{2 b}{g}} \Rightarrow \frac{t_{a}}{t_{b}} = \sqrt{\frac{a}{b}}$