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Standard XII

Physics

Shear Stress

A rod of leng...

Question

A rod of length 50 cm and area of cross section $4 m m^{2}$ is pulled at both ends by force of magnitude 100 N. The longitudinal stress in the rod is

2.5 \times 10^{7}

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2.5 \times 10^{6}

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200 Pa

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400 Pa

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Solution

The correct option is A $2.5 \times 10^{7}$ Pa
$\begin{matrix} Given length of rod = 50 c m Area of cross section = 4 {m m}^{2} Force = 100 N Stress = \frac{Force}{Area} = \frac{100}{4 \times 10^{- 6}} Stress = 2.5 \times 10^{7} P a \end{matrix}$

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Similar questions

Q. A copper rod with length

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[Y_{s t e e l} = 2 \times 10^{11} P a; Y_{C U} = 1.1 \times 10^{11} P a]

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A brass rod of length 2 m and cross-sectional area $2 {cm}^{2}$ is attached end to end to a steel rod of length L and cross-sectional area $1 {cm}^{2}$ . The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^{4} N$ at its ends. If the elongations of the two rods are equal, then the length (in m) the steel rod(L) is___

$[Y_{B} = 1 \times 10^{11} N / m^{2}$ and $Y_{S} = 2 \times 10^{11} N / m^{2}]$

Q. A brass rod of length 2 m and cross-sectional area 2.0

c m^{2}

is attached end to end to a steel rod of length L and cross-sectional area 1.0

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The compound rod is subjected to equal and opposite pulls of magnitude

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Y_{B r a s s} = 1.0 \times 10^{11} N / m^{2} a n d Y_{S t e e l} = 2.0 \times 10^{11} N / m^{2}

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A rod of length 50 cm and area of cross section 4 mm2 is pulled at both ends by force of magnitude 100 N. The longitudinal stress in the rod is

The correct option is A 2.5×107 Pa Given length of rod =50 cm Area of cross section =4 mm2 Force =100 N Stress = Force Area =1004×10−6Stress =2.5×107 Pa

A rod of length 50 cm and area of cross section $4 m m^{2}$ is pulled at both ends by force of magnitude 100 N. The longitudinal stress in the rod is

The correct option is A $2.5 \times 10^{7}$ Pa
$\begin{matrix} Given length of rod = 50 c m Area of cross section = 4 {m m}^{2} Force = 100 N Stress = \frac{Force}{Area} = \frac{100}{4 \times 10^{- 6}} Stress = 2.5 \times 10^{7} P a \end{matrix}$