NEET
Practice Test 8
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The diagram shown below represents the applied for cesper unit area with the corresponding change X (per unitlength) produced in a thin wire of uniform cross section in the curve shown. The region in which the wire behaves like aliquid is

  • Solution

    The wire starts behaving like a liquid at point b. It behaves like a viscous liquid in the region bc of the graph.

Two wires of same material and length but cross-sections in the ratio 1 : 2 are used to suspend the same loads. The extensions in them will be in the ratio

  • Solution

    Let W newton be the load suspended. Then

    \(Y=\frac{(W/A_{1})}{(\iota _{1}/L)}=\frac{WL}{A_{1}\iota_{1} }\)...(1)

    and \(Y=\frac{(W/A_{2})}{\iota _{2}/L}=\frac{WL}{A_{2}\iota _{2}}\)...(2)

    Dividing equation (1) by equation (2), we get

    \(1=\left ( \frac{\iota _{2}}{\iota _{1}} \right )\left ( \frac{A_{2}}{A_{1}} \right )=\left ( \frac{\iota _{2}}{\iota _{1}} \right )\left ( \frac{2}{1} \right )\)

    \(∴\frac{\iota _{1}}{\iota _{2}}=\frac{2}{1}\: or\: \iota _{1}:\iota _{2}= 2: 1\)

A body of mass 10 kg is attached to a wire of radius 3 cm. It’s breaking stress is 4.8 × 107 Nm-2, the area of cross-section of the wire is 10-6 m2. What is the maximum angular velocity with which it can be rotated in the horizontal circle ?

  • Solution

    Given that F/A = 4.8 × 107 Nm-2

    ∴ F = 4.8 × 107 × A or

    \(\frac{mv^{2}}{r}=4.8\times 10^{7}\times 10^{-6}=48\)

    or \(\frac{mr^{2}\omega ^{2}}{r}=48\: or\: \omega ^{2}=\frac{48}{mr}\)

    \(\omega =\sqrt{\left ( \frac{48}{10\times 0.3} \right )}=\sqrt{16}=4\: rad/sec\)

The Young’s modulus of brass and steel are respectively 1010 N/m2. and 2 × 1010 N/m2. A brass wire and a steel wire of the same length are extended by 1mm under the same force,the radii of brass and steel wiresare RB and RS respectively. Then

  • Solution

    We know that Y = F L/π r2 π or r2= FL/(Y π l)

    \(∴R_{B}^{2}=FL/(Y_{B}\pi \iota )\: and\: R_{s}^{2}=FL/(Y_{s}\pi \iota )\)

    \(or\frac{R_{B}^{2}}{R_{s}^{2}}=\frac{Y_{s}}{Y_{B}}=\frac{2\times 10^{10}}{10^{10}}=2\)

    \(orR_{B}^{2}=2R_{s}^{2}\: or\: R_{B}=\sqrt{2}R_{s}\)

    ∴\(R_{s}=R_{s}/\sqrt{2}\)

A steel wire of length 20 cm and uniform cross-section 1 mm2 is tied rigidly at both the ends. The temperature of the wire is altered from 40°C to 20°C. Coefficient of linear expansion for steel a= 1.1 × 10-5/°C and Y for steel is 2.0 × 1011 N/m2. The change in tension of the wire is

  • Solution

    F= Y A α t = (2.0 × 1011) (10-6) (1.1 × 10-5) (20)
    = 44 newton

Two rods A and B of the same material and length have their radii r1 and r2 respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, the ratio \(\left ( \frac{Angel\: of\: twist\: at\: the\: end\: of\: A}{Angel\: of\: twist\: at\: the\: end\: of\: B} \right )\) is

  • Solution

    Couple per unit angle of twist, \(C=\frac{\pi \eta r^{4}}{2\iota }\)

    ∴ Couple τ=C θ=\(\frac{\pi \eta r^{4}\Theta }{2\iota }\)

    Here η, l, C & τ are same. So, r4θ= constant

    \(∴\frac{\Theta _{1}}{\Theta _{2}}=\left ( \frac{r_{2}^{4}}{r_{1}^{4}} \right )\)

The length of a metal is l1 when the tension init is T1 and is l2 when the tension is T2. The original length of the wire is

  • Solution

    If l is the original length of wire, then change in length of first wire, \(\Delta \iota _{1}=(\iota _{1}-\iota )\)

    change in length of second wire,\(\Delta \iota _{1}=(\iota _{1}-\iota )\)

    Now,Y=\(\frac{T_{1}}{A}\times \frac{\iota }{\Delta \iota _{1}}=\frac{T_{2}}{A}\times \frac{\iota }{\Delta \iota _{2}}\)

    or \(\frac{T_{1}}{A}\times \frac{\iota }{\Delta \iota _{1}}=\frac{T_{2}}{A}\times \frac{\iota }{\Delta \iota _{2}}\)

    or \(T_{1}\iota _{2}-T_{1}\iota =T_{2}\iota _{1}-\iota T_{2}\: or\: \iota =\frac{T_{2}\iota _{1}-T_{1\iota _{2}}}{T_{2}-T_{1}}\)

A metal rod of Young’s modulus 2 × 1010 N m-2 under goes an elastic strain of 0.06%. The energy per unit volume stored in J m-3 is

  • Solution

    U /volume =1⁄2Y×strain2= 3600 J m-3
    [Strain = 0.06 × 10-2]

Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules in side will

  • Solution

    The molecular kinetic energy increases, and so termperature increases.

A 2 m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radian. The shear strain developed will be

  • Solution

    ɸ = \(\frac{r\Theta }{\iota }=\frac{1}{100}\times \frac{0.8}{2}= 0.004\: radian\)

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MECHANICAL PROPERTIES OF SOLIDS
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