Question

A square shaped slab of lead with side 50 cm and thickness 10.0 cm is subjected to a shearing force (on its narrow face) of magnitude 9.0×104 N. The lower edge is riveted to the floor as shown in figure. How much is the upper edge displaced, if the shear modulus of lead is 5.6×109 Pa?

- 4.8×10−4 m
- 3.2×10−4 m
- 1.6×10−4 m
- 0.8×10−4 m

Solution

The correct option is **C** 1.6×10−4 m

Given,

L=50 cm=50×10−2 m

t=10 cm=10×10−2 m

G=5.6×109 Pa

F=9.0×104 N

Area of the face on which force is applied,

A=50×10=500 cm2=0.05 m2

Let ΔL be the displacement of the upper edge of the slab, due to tangential force F applied.

Then, G=(FA)(ΔLL)

⇒ΔL=FLGA=9×104×50×10−25.6×109×0.05

∴ΔL=1.6×10−4 m

Given,

L=50 cm=50×10−2 m

t=10 cm=10×10−2 m

G=5.6×109 Pa

F=9.0×104 N

Area of the face on which force is applied,

A=50×10=500 cm2=0.05 m2

Let ΔL be the displacement of the upper edge of the slab, due to tangential force F applied.

Then, G=(FA)(ΔLL)

⇒ΔL=FLGA=9×104×50×10−25.6×109×0.05

∴ΔL=1.6×10−4 m

Suggest corrections

0 Upvotes

Similar questions

View More...

People also searched for

View More...

- About Us
- Contact Us
- Investors
- Careers
- BYJU'S in Media
- Students Stories - The Learning Tree
- Faces of BYJU'S – Life at BYJU'S
- Social Initiative - Education for All
- BYJU'S APP
- FAQ
- Support

© 2021, BYJU'S. All rights reserved.