The diameter of a brass rod is 4 mm and Young's modulus of brass is $9 \times {10^{10}}\,N/{m^2}$. The force required to stretch by $0.1\%$ of its length is
Medium
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionThe following four wires are made of same material. Which one will have the largest elongation when subjected to the same tension ?
- 2When a $4\, kg$ mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by $2\, cms$. The work required to be done by an external agent in stretching this spring by $5\, cms$ will be ......... $joule$ $(g = 9.8\,metres/se{c^2})$View Solution
- 3View SolutionWhich is correct relation
- 4In the given figure, two elastic rods $A$ & $B$ are rigidly joined to end supports. $A$ small mass $‘m’$ is moving with velocity $v$ between the rods. All collisions are assumed to be elastic & the surface is given to be frictionless. The time period of small mass $‘m’$ will be : [$A=$ area of cross section, $Y =$ Young’s modulus, $L=$ length of each rod ; here, an elastic rod may be treated as a spring of spring constant $\frac{{YA}}{L}$ ]View Solution
- 5Two wires $A$ and $B $ of same length and of the same material have the respective radii ${r_1}$ and ${r_2}$. Their one end is fixed with a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of $A$ and the angle of twist at the end of $B $ will beView Solution
- 6Consider the situation shown in figure. The force $F$ is equal to the $m_2g/2.$ If the area of cross-section of the string is $A$ and its Young's modulus $Y$, find the strain developed in it. The string is light and there is no friction anywhereView Solution
- 7A fixed volume of iron is drawn into a wire of length $L.$ The extension $x$ produced in this wire by a constant force $F$ is proportional toView Solution
- 8View SolutionThe Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of
- 9View SolutionCorrect pair is ..........
- 10The bulk modulus of rubber is $9.1\times 10^8\,N/m^2$. To ......... $m$ depth a rubber ball be taken in a lake so that its volume is decreased by $0.1\,\%$ .View Solution