TIFR CSE 2017 | Part A | Question: 3

Question

On planet TIFR, the acceleration of an object due to gravity is half that on planet earth. An object on planet earth dropped from a height $h$ takes time $t$ to reach the ground. On planet TIFR, how much time would an object dropped from height $h$ take to reach the ground?

• A. $\left(\dfrac{t}{\sqrt{2}}\right)$
• B.  $\sqrt {2}t$
• C.   $2t$
• D. $\left(\dfrac{h}{t}\right)$
• E. $\left(\dfrac{h}{2t}\right)$

Comments

There's a typo in option B, it should be $\sqrt{2} \cdot t$ instead of $\sqrt{2t}$

Answer 1

Let, The acceleration due to gravity on earth =$g.$

and the acceleration due to gravity on $TIFR = G=\dfrac{g}{2}.$

Time taken to reach the ground on earth $= t = \sqrt{\dfrac{2h}{g}}$.

Similarly, on TIFR planet, time taken = T = $\sqrt{\dfrac{2h}{G}}$.= $\sqrt{\dfrac{4h}{g}}$.

$\Rightarrow T = \sqrt{2}t.$

Ans - Option (b)

Comments

TIFR asks questions on Physics :)

Answer 2

Using basic laws of kinematics :

h = 1/2 a t²   for an object relaesed from top which is initially at rest

So a = g for earth we can write :

t₁  =  sqrt (2h / g)

Now according to the question

acceleration due to gravity in TIFR planet =  g / 2

So t₂  = sqrt(4h / g)  = sqrt(2) *  sqrt (2h / g)

                              = sqrt(2) * t₁ ..

Hence B) should be corrected answer..Only sqrt(2) should be there and 't' should be separate..Hence sqrt(2) * t is correct option..