A lift starting from rest with a constant upward acceleration moves 1.5 m in 004 s. If a person standing
in the lift holds a packet of 2 kg by a string, then the tension in the string due to motion is
[UP CPMT 2013]

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

57.1 N

67.1 N

5.89 N

None of these

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given, u = 0, s = 1.5 m, t = 0.4 s

From second equation of motion, $s = u t + \frac{1}{2} a t^{2}$

$\Rightarrow 1.5 = 0 + \frac{1}{2} a (0.4)^{2}$

$\Rightarrow a = \frac{1.5 \times 2}{(0.4)^{2}} = 18.75 {ms}^{- 2}$

As the string is moving upwards with this acceleration

$∴ T = m (g + a) = 2 (9.8 + 18.75) = 57.1 N$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test