Tag: properties of material substances

Quizzes (4) Questions (125)

Questions Related to properties of material substances

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

In an experiment on the determination of Young's Modulus of a wire by Searle's method, following data is available:
Normal length of the wire (L) = $110$cm
Diameter of the wire (d) = $0.01cm$
Elongation in the wire(l) = $0.125cm$
This elongation is for a tension of $50$N. The least counts for corresponding quantities are $0.01cm, 0.00005 cm, $ and $0.001cm$, respectively. Calculate the maximum error in calculating the value of Young's modulus(Y).

  1. $8\%$

  2. $1.809\%$

  3. $1.09\%$

  4. cant say
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Y = (F*L) / (A*l) = (4*F*L) / (pi*d^2*l). The relative error is deltaY/Y = deltaF/F + deltaL/L + 2*deltad/d + deltal/l. Plugging in the values: 0.01/50 (negligible) + 0.01/110 + 2*(0.00005/0.01) + 0.001/0.125. Calculation yields approximately 1.809%.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

When a weight of 5 kg is suspended from a copper wire of length 30 m and diameter 0.5 mm, the length of the wire increases by 2.4 cm. If the diameter is doubled, the extension produced is :

  1. 1.2 cm

  2. 0.6

  3. 0.3 cm

  4. 0.15 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Y \, = \, \dfrac{Mg \, \times \, 4 \, \times \, 1}{\pi D^2 \, \times \, \Delta l} \, or \, \Delta l \, \propto \, \dfrac{1}{D^2}$
(i) when D is double, $\Delta l$ becomes one-fourth. i.e.,
$\dfrac{1}{4} \, \times \, 2.4 \, cm \, i.e., 0.6 \, cm \, \times \, 2.4 \, cm$ i.e. 0.6 cm.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

The maximum load a wire can with stand without breaking, when it is stretched to twice of its original length, will:

  1. be half

  2. be four time decreased

  3. be double

  4. remain same
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

the maximum load a wire can with stand without breaking it is stretched to twice of its original length is remain same$.$

Hence,
option $(D)$ is correct answer.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

A uniform wire of length L and radius r is twisted by a angle $ \angle \alpha$. If modulus of rigidity of the wire is $ \eta  $, then the elastic potential energy stored in wire, is

  1. $ \frac{\pi \eta r^{4}\alpha }{2L^{2}} $

  2. $ \frac{\pi \eta r^{4}\alpha^{2} }{4L} $

  3. $ \frac{\pi \eta r^{4}\alpha }{4L^{2}} $

  4. $ \frac{\pi \eta r^{4}\alpha^{2} }{2L} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The elastic potential energy stored in a twisted wire is given by the formula U = (1/2) * C * alpha^2, where C is the torsional rigidity (C = (pi * eta * r^4) / (2 * L)). Substituting C into the formula yields U = (pi * eta * r^4 * alpha^2) / (4 * L).

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

The length of an elastic string is $x$ metre when the tension is $8\ N$. Its length is $y$ metre when the tension is $10\ N$. What will be its length, when the tension is $18\ N$?

  1. $2x + y$

  2. $5y - 4x$

  3. $7y - 5x$

  4. $7y + 5x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let, original length of the spring is L metre and, Y = $\dfrac { F.L }{ A.l } $

Now, when F = 8N, and l = (x - l)m then, $Y=\dfrac { 8.L }{ A.\left( x-L \right)  } \quad \longrightarrow (I)$
and when F=10N, and l = (y - l)m then, $y=\dfrac { 10.L }{ A.\left( y-L \right)  } m\quad \longrightarrow (II)$
From equation (I) and (II) we get,
$\dfrac { 8L }{ A\left( x-L \right)  } =\dfrac { 10L }{ A\left( y-L \right)  } $
or,  $8\left( y-L \right) =10\left( x-L \right) $
or,    $4y-4L=5x-5L$
or,                $L=5x-4y$
When, F=18N,
Let, length of the wire will be Z metre.
$\therefore \quad Y=\dfrac { 18.L }{ A.\left( Z-L \right)  } \quad \longrightarrow (III)$
From equation (I) and (III) we get,
$\dfrac { 8L }{ A\left( x-L \right)  } =\dfrac { 18L }{ A\left( Z-L \right)  } $
or,  $9\left( x-L \right) =4\left( Z-L \right) $
or,  $4Z=9x-9L+4L$
            $=9x-5L$
            $=9x-25x+20y$    [putting value of L]
or,  $Z=5y-4x$


Multiple choice elastic energy properties of material substances elasticity properties of matter physics

Work done by restoring force in a string within elastic limit is $-10\ J$. The maximum amount of heat produced in the string is :

  1. $10\ J$

  2. $20\ J$

  3. $5\ J$

  4. $15\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The work done by a restoring force is negative when the string is stretched. The energy stored in the string is converted into heat when the string returns to its equilibrium position, equal in magnitude to the work done.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

If work done in stretching a wire by 1 mm is 2J. Then the work necessary for stretching another wire of same material but with double the radius and half the length by 1 mm in joule is

  1. 1/4

  2. 4

  3. 8

  4. 16
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The stretching force $F=\dfrac{YA\Delta l}{l}$

where $Y=$Young's modulus, $A=$Area of cross-section of wire, $l=$actual length of wire, $\Delta l=$increase in length.
$F=\dfrac{Y\pi r^{2}\Delta l}{l}$
As the material is same $Y$ does not change.
$\dfrac{F _1}{F _2}=\dfrac{\dfrac{r _1^{2}\Delta l _1}{l _1}}{\dfrac{r _2^{2}\Delta l _2}{l _2}}$
Here $\Delta l _1=1mm$
$\Delta l _2=1mm$
$l _2=\dfrac{1}{2}l _1$
$r _2=2r _1$
$\dfrac{F _1}{F _2}=\dfrac{\dfrac{r _1^{2}\times 1mm}{l _1}}{\dfrac{4r _1^{2}\times 1mm}{\dfrac{1}{2}l _1}}$
$\dfrac{F _1}{F _2}=\dfrac{1}{8}$
The work done in stretching wire by amount $\Delta l$ is $W=\dfrac{1}{2} F\Delta l$
Hence $\dfrac{W _1}{W _2}=\dfrac{F _1}{F _2}=\dfrac{1}{8}$
As $F _1=2$
$F _2=2\times 8=16$
Hence the correct option is (D).

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

When a body mass $M$ is attached to power end of a wire (of length $L$) whose upper end is fixed, then the elongation of the wire is $l$. In this situation mark out the correct statement(s).

  1. Loss in gravitational potential energy of $M$ is $Mgl$.

  2. Elastic potential energy stored in the wire is $\dfrac {Mgl}{2}$

  3. Elastic potential energy stored in the wire is $Mgl$

  4. Elastic potential energy stored in the wire is $\dfrac {Mgl}{3}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Since it moves $l$ distance against gravity, so gravitational potential energy=Mgl
Elastic potential energy=$1/2\times Stress\times Strain\times Volume=1/2\times \dfrac{Mg}{A} \times \dfrac{l}{L}\times AL=\dfrac{Mgl}{2}$

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

Which of the following statements is correct regarding Poisson's ratio?

  1. It is the ratio of the longitudinal strain to the lateral strain

  2. Its value is independent of the nature of the material

  3. It is unitless and dimensionless quantity

  4. The practical value of Poisson's ratio lies between $0$ and $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of the lateral strain to longitudinal strain is called Poisson's ratio.
Hence, option (a) is an incorrect statement.
Its value depends only on the nature of the material. 
Hence, option (b) is an incorrect statement.
It is the ratio of two like physical quantities.
Therefore, it is unitless and dimensionless quantity.
Hence option (c) is a correct statement
The practical value of Poisson's ratio lies between $0$ and $0.5$
hence option (d) is an incorrect statement.