Ncert Exercises & Solutions

A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (figure). The cross-sectional areas of wires A and B are $1.0 {mm}^{2} and 2.0 {mm}^{2}$ respectively. ( $Y_{Al} = 70 \times 10^{9} N / m^{2} and Y_{steel} = 200 \times 10^{9} N / m^{2}$ )

a.	Mass m should be suspended close to wire A to have equal stresses in both wires.
b.	Mass m should be suspended close to B to have equal stresses in both wires.
c.	Mass m should be suspended in the middle of the wires to have equal stresses in both wires.
d.	Mass m should be suspended close to wire A to have equal strain in both wires.

The correct statements are:

1. (b, c)

2. (a, d)

3. (b, d)

4. (c, d)

(3) Hint: The stress and strain in the wires depend on Young's modulus of the wires.

Step 1: Find the stresses in the wires.

Let the mass is placed at x from end B.

Let

T_{A}

and

T_{B}

be the tensions in wire A and wire B respectively.

For the rotational equilibrium of the system,

Τ_{ζ} = 0

(Total torque = 0)

\Rightarrow

T_{B} x - T_{A} (l - x) = 0

\Rightarrow

\frac{T_{B}}{T_{A}} = \frac{l - x}{x}

...(i)

Stress in wire A,

= S_{A} = \frac{T_{A}}{a_{A}}

Stress in wire B,

S_{B} = \frac{T_{B}}{a_{B_{}}}

where

a_{A} and a_{B}

are cross-sectional areas of wires A and B respectively.

Step 2: Find the location of the mass for equal stresses in the wires.

According to the question,

a_{B} = 2 a_{A}

Now, for equal stresses,

S_{A} = S_{B}

\Rightarrow \frac{T_{A}}{a_{A}} = \frac{T_{B}}{a_{B}} \Rightarrow \frac{T_{B}}{T_{A}} = \frac{a_{B}}{a_{A}} = 2

\Rightarrow \frac{l - x}{x} = 2 \Rightarrow \frac{l}{x} - 1 = 2

\Rightarrow x = \frac{l}{3} \Rightarrow l - x = l - l / 3 = \frac{2 l}{3}

Hence, the mass m should be placed closer to B.

Step 3: Find the location of the mass for equal strain in the wires.

For equal strain,

{(strain)}_{A} = {(strain)}_{B}

\Rightarrow \frac{(Y_{A})}{S_{A}} = \frac{Y_{B}}{S_{B}} (Where Y_{A} and Y_{B} are Young' s moduli)

\Rightarrow \frac{Y_{steel}}{T_{A} / a_{A}} = \frac{Y_{Al}}{T_{B} / a_{B}}

\Rightarrow \frac{Y_{steel}}{Y_{Al}} = \frac{T_{A}}{T_{B}} \times \frac{a_{B}}{a_{A}} = (\frac{x}{l - x}) (\frac{2 a_{A}}{a_{A}})

\Rightarrow \frac{200 \times 10^{9}}{70 \times 10^{9}} = \frac{2 x}{l - x} \Rightarrow \frac{20}{7} = \frac{2 x}{l - x}

\Rightarrow \frac{10}{7} = \frac{x}{l - x} \Rightarrow 10 l - 10 x = 7 x

\Rightarrow 17 x = 10 l \Rightarrow x = \frac{10 l}{17}

l - x = l - \frac{10 l}{17} = \frac{7 l}{17}

Hence, the mass m should be placed closer to wire A.

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