In 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 $\frac{{YA}}{L}$ ]
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.*
The upper end of a wire of radius $4\, mm$ and length $100\, cm$ is clamped and its other end is twisted through an angle of $60^o$. Then angle of shear is .......... $^o$
For a constant hydraulic stress on an object, the fractional change in the object's volume $\left( {\frac{{\Delta V}}{V}} \right)$ and its bulk modulus $(B)$ are related as
A uniform rod of length $L$ has a mass per unit length $\lambda$ and area of cross-section $A$. If the Young's modulus of the rod is $Y$. Then elongation in the rod due to its own weight is ...........
A solid cube of copper of edge $10 \,cm$ subjected to a hydraulic pressure of $7 \times 10^6\, pascal$. If Bulk modulus of copper is $140 \,GPa$, then contraction in its volume will be ................ $m ^3$
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
The proportional limit of steel is $8 \times 10^8 \,N / m ^2$ and its Young's modulus is $2 \times 10^{11} \,N / m ^2$. The maximum elongation, a one metre long steel wire can be given without exceeding the elastic limit is $......mm$
A wire of length $L$ and radius $r$ is rigidly fixed at one end. On stretching the other end of the wire with a force $F$, the increase in its length is $l$. If another wire of same material but of length $2L$ and radius $2r$ is stretched with a force of $2F$, the increase in its length will be