Tag: properties of matter

Questions Related to properties of matter

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


Correct Option: B

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


Correct Option: B
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.

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


Correct Option: D
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.

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} $


Correct Option: B

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. $7y + 5x$


Correct Option: B
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,  $Z=5y-4x$


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. $15\ J$


Correct Option: A

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


Correct Option: D
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).

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 $\dfrac {Mgl}{3}$


Correct Option: A,B
Explanation:

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

Viscosity is most closely related to:

  1. density

  2. velocity

  3. friction

  4. energy


Correct Option: C
Explanation:

Viscosity is the property by virtue of which a liquid offers resistance for the relative motion between liquid molecules. Friction is the force which opposes relative motion between any two objects. So, viscosity is most closely related to friction.

Viscosity of the fluids is analogous to:

  1. Random motion of the gas molecules

  2. Friction between the solid surfaces

  3. integral motion

  4. Nonuniform motion of solids


Correct Option: B
Explanation:

Viscosity is the property by virtue of which a liquid offers resistance for the relative motion between its layer.
Friction is the force between two solid surfaces which offers resistance for the relative motion between them. 
So, viscosity of the fluids is analogous to friction between the solid surfaces.