JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 14)
An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s$$^2$$)
_25th_January_Morning_Shift_en_14_1.png)
90 N
240 N
30 N
300 N
Explanation
_25th_January_Morning_Shift_en_14_2.png)
$\frac{T}{2} \times 0.6=20 \times 0.5+80 \times 1$
$T \times 0.3=10+80=90$
$T=\frac{900}{3}=300 \mathrm{~N}$
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