5. A bar of length 100 cm' is supported at its two ends. The breadth and depth of bar are 5 cm and 0.5 cm respectively. A mass of 100 g is suspended at the centre of bar. Compute the depression produced in the bar. (Y = 4 x 10¹⁰ N/m²) (Ans. : 9.8 x 10⁻⁴ m)
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Given :
l = 100 cm = 1 m,
b = 5 cm = 5 × 10⁻² m,
d = 0.5 cm = 0.5 × 10⁻² m,
m = 100g = 0.1 kg,
Y = 4 × 10¹⁰N/m²
δ = Wl³/ 4Ybd ³
=0.1x9.8 x (1)³/4x4x10¹⁰x5x10⁻² x(0.5 x 10⁻²)³
=9.8x10⁻⁴m
∴The depression produced in the bar is 9.8x10⁻⁴m
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Answer:
Explanation:
Given :
l = 100 cm = 1 m,
b = 5 cm = 5 × 10⁻² m,
d = 0.5 cm = 0.5 × 10⁻² m,
m = 100g = 0.1 kg,
Y = 4 × 10¹⁰N/m²
δ = Wl³/ 4Ybd ³
=0.1x9.8 x (1)³/4x4x10¹⁰x5x10⁻² x(0.5 x 10⁻²)³
=9.8x10⁻⁴m
∴The depression produced in the bar is 9.8x10⁻⁴m
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