Question

48. On two knife-edges separated by a distance of 0.6 m, a cylindrical beam of length 1 mis placed horizontally
and symmetrically. When a load of 7.3 kg is suspended from each end, the elevation of the centre is
4.75 x 10 m. Find the Young's modulus of the beam. The radius of the beam - 7* 10-m.
(Ans. 7.23 x 100 Nm ?​

Answer 1

✓ Correct question

On two knife - edges separated by a distance of 0.6 m, a cylindrically beam of length 1m is placed horizontally and symmetrically. When a load of 7.3 Kg is suspended from each end , the elevation of the centre is 4.75 * 10^-3 m. Find the young's modulus of the beam. The radius of the beam = 7 * 10 ^-3m .

✓ Given

» Distance separated on two knife - edges = 0.6 m

» Cylindrical beam length = 1m

» Mass of the load = 7.3 Kg

» Elevation of the centre = 4.75 * 10^ -3 m

» radius of the beam = 7 * 10 ^-3m

We know that g = 9.8 m/s^2

✓ To find

Young' modulus ( E ) of the beam

✓ Formula used

$\bf \: e = \frac{mga {l}^{2} }{2b {d}^{3y} } \: \: n {m}^{ - 2}$

✓ Solution

$\bf \: e = \frac{mga {l}^{2} }{2b {d}^{3y} } \: \: n {m}^{ - 2}$

By substituting all the values in the above formula , we get

$\frac{3 \times 7.3 \times 9.8 \times 20 \times 0.6 \times 0.6}{2 \times 0.02 \times { (14 \times {10}^{ - 3})}^{3} \times 4.75 \times {10}^{ - 3} }$

$= \frac{4292.4 \times 0.36}{0.19 \times 2744 \times {10}^{ - 9} \times {10}^{ - 3} }$

$= \frac{1545.26}{521.36 \times {10}^{ - 12} }$

$= 2.96 \times {10}^{12}$

Therefore the young' modulus (E) = 2.96× 10^ 12 Nm^-2

______________________

An question arises that thickness (d) of the beam is not given here , Yes !! It's not been provided .

Usually when we do it in virtual lab , We measure the diameter of the beam using screw gauge ( so here the radius of the beam is given !! )

Diameter = 2r

$diameter = 2 \times 7 \times {10}^{ - 3} = 14 \times {10}^{ - 3 \: } m$

Here the thickness ( d ) is the diameter's value .

And We took that a = 20

what's a here ??

a is the distance between the scale and the material which is hung from it ( here it's load of 7.3 Kg)

The scale usually is 100 cm length , in order to hung it we need to remove 60 cm ( given in question as 0.6 m ) [ Uniform bending ]

______________________