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Laws of Motion

JEE Main level, numerical answer, medium

163 of 210

Question 163 of 210

Two blocks ( $m=0.5kg$ and $M=4.5kg$ ) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is $\dfrac{3}{7}.$ Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is $N.$ (Round off to the Nearest Integer) [Take $g$ as $9.8m{s}^{-2}$ ]

Show answer and solution

Answer: 21

{a}_{\max }=\mu g=\dfrac{3}{7}\times 9.8

F=(M+m){a}_{\max }=5{a}_{\max }

=21N

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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