Two wires are arranged as shown in the figure. The elongations in upper and lower wires are respectively

\frac{(m_{1} + m_{2}) g ℓ_{1}}{A Y_{1}}, \frac{m_{2} g ℓ_{2}}{A Y_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{(m_{1} - m_{2}) g ℓ_{1}}{A Y_{1}}, \frac{m_{2} g ℓ_{2}}{A Y_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(\frac{m_1}{m_2} + 1) g ℓ_1}{A Y_1}, \frac{m_2 g ℓ_2}{A Y_2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(\frac{m_{1}}{m_{2}} - 1) g ℓ_{1}}{A Y_{1}}, \frac{m_{2} g ℓ_{2}}{A Y_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{(m_{1} + m_{2}) g ℓ_{1}}{A Y_{1}}, \frac{m_{2} g ℓ_{2}}{A Y_{2}}$
$Let Tension in upper and Lower wire be T_{1}, T_{2} respectlively.$

$T_{2} = m_{2} g$

$\begin{matrix} T_{1} = T_{2} + m_{1} g T_{1} = (m_{1} + m_{2}) g \end{matrix} (since T_{2} = m_{2} g)$

$we know , elongation, Δ l = \frac{F l}{A y}$

$Δ l_{1} = \frac{(m_{1} + m_{2}) g L_{1}}{A_{1} Y_{1}}$

$Δ l_{2} = \frac{m_{2} g l_{2}}{A_{2} Y_{2}}$

Suggest Corrections

Similar questions

Q. As shown in adjacent figure it a load of mass (m) is attached at lower end of wire. Then find the displacement of the point B,C and D are as shown in figure,
(i) elongation of first wire

e_{1} = \frac{(m g) ℓ_{1}}{A Y_{1}}

(ii) elongation of 2nd wire

e_{2} = \frac{(m g) ℓ_{2}}{A Y_{2}} + \frac{(m g) ℓ_{1}}{A Y_{1}}

(iii) elongation of 3rd wire

e_{2} = \frac{(m g) ℓ_{2}}{A Y_{3}} + \frac{(m g) ℓ_{1}}{A Y_{2}} + \frac{(m g) ℓ_{1}}{A Y_{1}}

Q. Two blocks of masses

m_{1}

and

m_{2}

are connected as shown in the figure. The acceleration of the block

m_{2}

Q. In the arrangement, shown in figure, pulleys are massless and frictionless and threads are inextensible. Blocks of mass m1 will remain at rest if

m₁ and m₂ are the slope of two perpendicular lines, if

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:

Join BYJU'S Learning Program

Two wires are arranged as shown in the figure. The elongations in upper and lower wires are respectively

The correct option is A (m1+m2)gℓ1AY1,m2gℓ2AY2 Let Tension in upper and Lower wire be T1,T2 respectlively. T2=m2gT1=T2+m1gT1=(m1+m2)g (since T2=m2g)we know , elongation, Δl=FlAyΔl1=(m1+m2)gL1A1Y1Δl2=m2gl2A2Y2