A force of ${10^3}$ newton stretches the length of a hanging wire by $1$ millimetre. The force required to stretch a wire of same material and length but having four times the diameter by $1$ millimetre is
A$4 \times {10^3}N$
B$16 \times {10^3}N$
C$\frac{1}{4} \times {10^3}N$
D$\frac{1}{{16}} \times {10^3}N$
Medium
Download our app for free and get started
B$16 \times {10^3}N$
b (b) $F = Y \times A \times \frac{l}{L}$ $⇒$ $F \propto {r^2}$ $(Y,l$ and $L$ are constant$)$
If diameter is made four times then force required will be $16$ times. i.e. $16 \times 10^3 N$
Download our app
and 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.*
A metal wire of length $0.5 m$ and cross-sectional area $10^{-4}\; m ^{2}$ has breaking stress $5 \times 10^{8}\; Nm ^{-2}$. A block of $10\; kg$ is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be $ms ^{-1}$.
Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^{3} {kg}$, The inner and outer radii of each column are $50\; {cm}$ and $100 \;{cm}$ respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use $\left.{Y}=2.0 \times 10^{11} \;{Pa}, {g}=9.8\; {m} / {s}^{2}\right]$
A thin $1 \,m$ long rod has a radius of $5\, mm$. A force of $50\,\pi kN$ is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is $0.01\, mm$, which of the following statements is false ?
An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is
A spherical ball contracts in volume by $0.02 \%$, when subjected to a normal uniform pressure of $50$ atmosphere. The Bulk modulus of its material is ............... $N / m ^2$