If the length of a wire is made double and radius is halved of its respective values. Then, the Young's modules of the material of the wire will :
If the length of a wire is made double and radius is halved of its respective values. Then, the Young's modules of the material of the wire will :
- [JEE MAIN 2022]
- A
Remains same
- B
Become $8$ times its initial value
- C
Become $\frac{1^{\text {th }}}{4}$ of its initial value
- D
Become $4$ times its initial value
Similar Questions
What is Young’s modulus ? Explain. and Give its unit and dimensional formula.
What is Young’s modulus ? Explain. and Give its unit and dimensional formula.
A mild steel wire of length $1.0 \;m$ and cross-sectional area $0.50 \times 10^{-2} \;cm ^{2}$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of $100 \;g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
A mild steel wire of length $1.0 \;m$ and cross-sectional area $0.50 \times 10^{-2} \;cm ^{2}$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of $100 \;g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
To double the length of a iron wire having $0.5\,c{m^2}$ area of cross-section, the required force will be $(Y = {10^{12}}\,dyne/c{m^2})$
To double the length of a iron wire having $0.5\,c{m^2}$ area of cross-section, the required force will be $(Y = {10^{12}}\,dyne/c{m^2})$
A stress of $1.5\,kg.wt/mm^2$ is applied to a wire of Young's modulus $5 \times 10^{11}\,N/m^2$ . The percentage increase in its length is
A stress of $1.5\,kg.wt/mm^2$ is applied to a wire of Young's modulus $5 \times 10^{11}\,N/m^2$ . The percentage increase in its length is
A rod of length $1.05\; m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Figure. The cross-sectional areas of wires $A$ and $B$ are $1.0\; mm ^{2}$ and $2.0\; mm ^{2}$. respectively. At what point along the rod should a mass $m$ be suspended in order to produce $(a)$ equal stresses and $(b)$ equal strains in both steel and alumintum wires.
A rod of length $1.05\; m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Figure. The cross-sectional areas of wires $A$ and $B$ are $1.0\; mm ^{2}$ and $2.0\; mm ^{2}$. respectively. At what point along the rod should a mass $m$ be suspended in order to produce $(a)$ equal stresses and $(b)$ equal strains in both steel and alumintum wires.