Question

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)

Answer 1

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

Answer 2

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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