A screw jack of pitch $5$ mm is used to lift the tyre of a vehicle of load $200$ kg with the help of a handle of length $0.5$ m. Neglecting the friction force between screw and nut of the jack, the least force required to raise the load is:

1.2 N

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2.2 N

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3.2 N

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4.2 N

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Solution

The correct option is A 3.2 N
Work done in one rotation $=$ work done against gravity
i.e. $F \times 2 π \times \frac{1}{2} = 200 \times 10 \times 5 \times 10^{- 3}$

$F \times 2 π = (m g) \times p i t c h$

$F = \frac{10}{3.14} = 3.18 N$

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A screw jack of pitch 5 mm is used to lift the tyre of a vehicle of load 200 kg with the help of a handle of length 0.5 m. Neglecting the friction force between screw and nut of the jack, the least force required to raise the load is:

The correct option is A 3.2 NWork done in one rotation = work done against gravityi.e. F×2π×12=200×10×5×10−3F×2π=(mg)×pitchF=103.14=3.18N

A screw jack of pitch $5$ mm is used to lift the tyre of a vehicle of load $200$ kg with the help of a handle of length $0.5$ m. Neglecting the friction force between screw and nut of the jack, the least force required to raise the load is:

The correct option is A 3.2 N
Work done in one rotation $=$ work done against gravity
i.e. $F \times 2 π \times \frac{1}{2} = 200 \times 10 \times 5 \times 10^{- 3}$

$F \times 2 π = (m g) \times p i t c h$

$F = \frac{10}{3.14} = 3.18 N$