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Physics · Current Electricity

JEE Main 2026 — 24 January, Morning Shift — Question 46

In the given figure the blocks A,B\mathrm{A}, \mathrm{B} and C weigh 4 kg,6 kg4 \mathrm{~kg}, 6 \mathrm{~kg} and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5 . The force F→\overrightarrow{\mathrm{F}} required to slide the block C with constant speed is ____\_\_\_\_ N . (Used g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} )

Question figure

Answer: 210

Numerical answer — enter this value.

Step-by-step solution

For 8 kg to move with constant velocity Fnet =0\mathrm{F}_{\text {net }}=0. ∴F=90+T+50\therefore \mathrm{F}=90+\mathrm{T}+50 (for 8kg block) T=20+50\mathrm{T}=20+50 (for 6 kg block) ∴F=210 N\therefore \mathrm{F}=210 \mathrm{~N}.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Current Electricity
Topic
Electrical Measuring Instruments

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