A uniform disc with mass \(\mathrm{M}=4 \mathrm{~kg}\) and radius \(\mathrm{R}=10 \mathrm{~cm}\) is mounted on a fixed horizontal axle as shown in figure. A block with mass \(\mathrm{m}=2 \mathrm{~kg}\) hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ______ N. (Take \(\mathrm{g}=10 \mathrm{~ms}^{-2}\))

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Rotational Mechanics

A uniform disc with mass

M = 4 kg

and radius

R = 10 cm

is mounted on a fixed horizontal axle as shown in figure. A block with mass

m = 2 kg

hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ______ N.
(Take

g = 10 ms^{- 2}

)

Answer: 10

Solution

For the falling block:

m g - T = ma

. For the rotating disc:

T R = I α = \frac{M R ^{2}}{2} α

. Since the cord doesn't slip,

a = R α

, so

α = \frac{a}{R}

. Substituting:

T = \frac{M R ^{2}}{2} \cdot \frac{a}{R ^{2}} = \frac{M a}{2}

. From the block equation:

m g - T = ma

, substituting

T = \frac{M a}{2}

m g = ma + \frac{M a}{2} = a (m + \frac{M}{2})

. Thus

a = \frac{m g}{m + \frac{M}{2}} = \frac{2 \times 10}{2 + 2} = 5 m/s^{2}

. Therefore

T = \frac{4 \times 5}{2} = 10 N

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Rotational Mechanics

Solution

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