Start free

Physics · Mechanical Properties of Matter

JEE Main 2026 — 5 April, Morning Shift — Question 19

A cube has side length 55 cm and modulus of rigidity 10510^5 N/m². The displacement produced by a force of 1010 N in the upper face of cube is ______ mm.

Question figure

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Shear modulus η=F/Aϕ=FhAx\eta = \frac{F/A}{\phi} = \frac{Fh}{A x} ⇒ x=FhAη=10×0.05(0.05)2×105=0.50.0025×105=0.5250=0.002x = \frac{Fh}{A\eta} = \frac{10 \times 0.05}{(0.05)^2 \times 10^5} = \frac{0.5}{0.0025 \times 10^5} = \frac{0.5}{250} = 0.002 m = 2 mm.

Answer key and solution verified before publishing.

Practise Mechanical Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Stress,Strain and Modulus of Elasticity

More Mechanical Properties of Matter questions from this paper

A cube has side length 5 cm and modulus of rigidity 10 5 N/m².… | JEE Main 2026 PYQ with Solution · DhiX AI