In $CGS$ system, the Young's modulus of a steel wire is $2 \times {10^{12}}$. To double the length of a wire of unit cross-section area, the force required is
Easy
Download our app for free and get started
(b)To double the length of wire,
Stress = Young's modulus
$\frac{F}{A} = 2 \times {10^{12}}\frac{{dyne}}{{c{m^2}}}.$
If $A = 1$ then $F = 2 × 10^{12}$ dyne
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
- 1A wire suspended vertically from one of its ends is stretched by attaching a weight of $200\, N$ to the lower end. The weight stretches the wire by $1\, mm$ Then the elastic energy stored in the wire is ........ $J$View Solution
- 2View 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
- 3View SolutionThe bulk modulus of an ideal gas at constant temperature
- 4A steel rod has a radius of $20\,mm$ and a length of $2.0\,m$. A force of $62.8\,kN$ stretches it along its length. Young's modulus of steel is $2.0 \times 10^{11}\,N / m ^2$. The longitudinal strain produced in the wire is $..........\times 10^{-5}$View Solution
- 5A glass slab is subjected to a pressure of $10\, atm$. The fractional change in its volume is (Bulk modulus of glass $= 37 \times 10^9\, N\, m^{-2}$, $1\, atm = 1 \times 10^5\, N\, m^{-2}$)View Solution
- 6Work done by restoring force in a string within elastic limit is $-10 \,J$. Maximum amount of heat produced in the string is .......... $J$View Solution
- 7A rigid massless rod of length $6\ L$ is suspended horizontally by means of two elasticrods $PQ$ and $RS$ as given figure. Their area of cross section, young's modulus and lengths are mentioned in figure. Find deflection of end $S$ in equilibrium state. Free end of rigid rod is pushed down by a constant force . $A$ is area of cross section, $Y$ is young's modulus of elasticityView Solution
- 8A load $W$ produces an extension of $1mm$ in a thread of radius $r.$ Now if the load is made $4W$ and radius is made $2r$ all other things remaining same, the extension will become..... $mm$View Solution
- 9Young's modulus of rubber is ${10^4}\,N/{m^2}$ and area of cross-section is $2\,c{m^2}$. If force of $2 \times {10^5}$ dynes is applied along its length, then its initial length $l$ becomesView Solution
- 10In the below graph, point $D$ indicatesView Solution
