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Standard XII

Physics

Thermal Expansion

A string of l...

Question

A string of length $L$ and mass $M$ is lying on horizontal table.A force $F$ is applied at one of its end. Tension in the string at a distance $x$ from the ends at which force is applied is

Zero

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F

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F (L - x) / L

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F (L - x) / M

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Solution

The correct option is C $F (L - x) / L$
Tension at distance $x$ , means tension on , a imaginary string at $n$ distance connecting the rope.
$F - T = (\frac{M}{L} \times n) \times a \frac{F}{M}$
$T = F - \frac{M x}{L} \times \frac{F}{M} = F (1 - \frac{x}{L}) T = \frac{F (L - x)}{L}$

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A string of length L and mass M is lying on horizontal table.A force F is applied at one of its end. Tension in the string at a distance x from the ends at which force is applied is

The correct option is C F(L−x)/LTension at distance x, means tension on , a imaginary string at n distance connecting the rope.F−T=(ML×n)×aFMT=F−MxL×FM=F(1−xL)T=F(L−x)L

A string of length $L$ and mass $M$ is lying on horizontal table.A force $F$ is applied at one of its end. Tension in the string at a distance $x$ from the ends at which force is applied is

The correct option is C $F (L - x) / L$
Tension at distance $x$ , means tension on , a imaginary string at $n$ distance connecting the rope.
$F - T = (\frac{M}{L} \times n) \times a \frac{F}{M}$
$T = F - \frac{M x}{L} \times \frac{F}{M} = F (1 - \frac{x}{L}) T = \frac{F (L - x)}{L}$