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PHXI06:WORK ENERGY AND POWER

355619 A spring of spring constant $2 N c m^{- 1}$ has an extension of 5 $c m$ . The minimum work done in joule in increasing the extension from 5 $c m$ to 15 $c m$ is

1 2

J

2 16

J

3 32

J

4 4

J

Explanation:

Here, $k = 2 N c m^{- 1} = 2 \times 10^{2} N m^{- 1}$
$x_{1} = 5 c m = 5 \times 10^{- 2} m$
$x_{2} = 15 c m = 15 \times 10^{- 2} m$
Work done $=$ Increase in PE of spring
$= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 2 \times 10^{2} [{(15 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = 10^{2} \times 200 \times 10^{- 4} = 2 J$
$= 10^{2} \times 200 \times 10^{- 4} = 2 J$

PHXI06:WORK ENERGY AND POWER

355620 Assertion :
A spring of force constant $k$ is cut into two pieces having lengths in the ratio $1 : 2$ . The force constant of series combination of the two parts is $3 k / 2$ .
Reason :
The spring connected in series are represented by $k = k_{1} + k_{2}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

As we know, $k = \frac{F}{l} \Rightarrow k \propto \frac{1}{l}$
$\Rightarrow \frac{k_{2}}{k_{1}} = \frac{l_{1}}{l_{2}} = \frac{1}{2}$
$k_{1} = 2 k, k_{2} = k$
In series, $\frac{1}{k^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} = \frac{1}{2 k} + \frac{1}{k} = \frac{3}{2 k}$
$k^{'} = \frac{2 k}{3}$
So, both Assertion and Reason are incorrect.
So Correct option is (4)

AIIMS - 2017

PHXI06:WORK ENERGY AND POWER

355621 Two blocks of mass 3 $k g$ and 6 $k g$ respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant $k = 200 N / m$ . Initially the spring is unstretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be :-
supporting img

1 30

c m

2 25

c m

3 20

c m

4 15

c m

Explanation:

We know that
$K . E_{s y s} = K . E_{o f C . M} + K . E_{w . r . t o C . M .}$
$K . E_{s y s} = \frac{1}{2} \times 3 \times 1^{2} + \frac{1}{2} \times 6 \times 2^{2} = 13.5 J$
$V_{C M} = \frac{6 \times 2 - 3 \times 1}{9} = 1 m / s$
$K . E_{o f C M} = \frac{1}{2} \times 9 \times 1^{2} = 4.5 J$
$\Rightarrow K . E_{w . r . t o C . M} = (13.5 - 4.5) J = 9 J$
In the absence of external force on a system
$K . E_{o f C M}$ - always remains constant
So $K . E_{w . r . t o c M}$ will be converted into $P . E$
$K . E_{w . r . t o C . M} = 9 J = \frac{1}{2} K x_{max}^{2}$
$x_{max} = 0.3 m = 30 c m$

PHXI06:WORK ENERGY AND POWER

355622 Two springs have their force constants $k_{1}$ and $k_{2} (K_{2} > K_{1})$ . When they are stretched by the same force, work done is

1 Same in both the springs

2 More in spring

k_{1}

3 More in spring

k_{2}

4 Data incorrect

Explanation:

$W = \frac{1}{2} k x^{2} = \frac{F^{2}}{2 k} \Rightarrow W α \frac{1}{k} (F =$ constant $)$

PHXI06:WORK ENERGY AND POWER

355619 A spring of spring constant $2 N c m^{- 1}$ has an extension of 5 $c m$ . The minimum work done in joule in increasing the extension from 5 $c m$ to 15 $c m$ is

1 2

J

2 16

J

3 32

J

4 4

J

Explanation:

Here, $k = 2 N c m^{- 1} = 2 \times 10^{2} N m^{- 1}$
$x_{1} = 5 c m = 5 \times 10^{- 2} m$
$x_{2} = 15 c m = 15 \times 10^{- 2} m$
Work done $=$ Increase in PE of spring
$= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 2 \times 10^{2} [{(15 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = 10^{2} \times 200 \times 10^{- 4} = 2 J$
$= 10^{2} \times 200 \times 10^{- 4} = 2 J$

PHXI06:WORK ENERGY AND POWER

355620 Assertion :
A spring of force constant $k$ is cut into two pieces having lengths in the ratio $1 : 2$ . The force constant of series combination of the two parts is $3 k / 2$ .
Reason :
The spring connected in series are represented by $k = k_{1} + k_{2}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

As we know, $k = \frac{F}{l} \Rightarrow k \propto \frac{1}{l}$
$\Rightarrow \frac{k_{2}}{k_{1}} = \frac{l_{1}}{l_{2}} = \frac{1}{2}$
$k_{1} = 2 k, k_{2} = k$
In series, $\frac{1}{k^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} = \frac{1}{2 k} + \frac{1}{k} = \frac{3}{2 k}$
$k^{'} = \frac{2 k}{3}$
So, both Assertion and Reason are incorrect.
So Correct option is (4)

AIIMS - 2017

PHXI06:WORK ENERGY AND POWER

355621 Two blocks of mass 3 $k g$ and 6 $k g$ respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant $k = 200 N / m$ . Initially the spring is unstretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be :-
supporting img

1 30

c m

2 25

c m

3 20

c m

4 15

c m

Explanation:

We know that
$K . E_{s y s} = K . E_{o f C . M} + K . E_{w . r . t o C . M .}$
$K . E_{s y s} = \frac{1}{2} \times 3 \times 1^{2} + \frac{1}{2} \times 6 \times 2^{2} = 13.5 J$
$V_{C M} = \frac{6 \times 2 - 3 \times 1}{9} = 1 m / s$
$K . E_{o f C M} = \frac{1}{2} \times 9 \times 1^{2} = 4.5 J$
$\Rightarrow K . E_{w . r . t o C . M} = (13.5 - 4.5) J = 9 J$
In the absence of external force on a system
$K . E_{o f C M}$ - always remains constant
So $K . E_{w . r . t o c M}$ will be converted into $P . E$
$K . E_{w . r . t o C . M} = 9 J = \frac{1}{2} K x_{max}^{2}$
$x_{max} = 0.3 m = 30 c m$

PHXI06:WORK ENERGY AND POWER

355622 Two springs have their force constants $k_{1}$ and $k_{2} (K_{2} > K_{1})$ . When they are stretched by the same force, work done is

1 Same in both the springs

2 More in spring

k_{1}

3 More in spring

k_{2}

4 Data incorrect

Explanation:

$W = \frac{1}{2} k x^{2} = \frac{F^{2}}{2 k} \Rightarrow W α \frac{1}{k} (F =$ constant $)$

PHXI06:WORK ENERGY AND POWER

355619 A spring of spring constant $2 N c m^{- 1}$ has an extension of 5 $c m$ . The minimum work done in joule in increasing the extension from 5 $c m$ to 15 $c m$ is

1 2

J

2 16

J

3 32

J

4 4

J

Explanation:

Here, $k = 2 N c m^{- 1} = 2 \times 10^{2} N m^{- 1}$
$x_{1} = 5 c m = 5 \times 10^{- 2} m$
$x_{2} = 15 c m = 15 \times 10^{- 2} m$
Work done $=$ Increase in PE of spring
$= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 2 \times 10^{2} [{(15 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = 10^{2} \times 200 \times 10^{- 4} = 2 J$
$= 10^{2} \times 200 \times 10^{- 4} = 2 J$

PHXI06:WORK ENERGY AND POWER

355620 Assertion :
A spring of force constant $k$ is cut into two pieces having lengths in the ratio $1 : 2$ . The force constant of series combination of the two parts is $3 k / 2$ .
Reason :
The spring connected in series are represented by $k = k_{1} + k_{2}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

As we know, $k = \frac{F}{l} \Rightarrow k \propto \frac{1}{l}$
$\Rightarrow \frac{k_{2}}{k_{1}} = \frac{l_{1}}{l_{2}} = \frac{1}{2}$
$k_{1} = 2 k, k_{2} = k$
In series, $\frac{1}{k^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} = \frac{1}{2 k} + \frac{1}{k} = \frac{3}{2 k}$
$k^{'} = \frac{2 k}{3}$
So, both Assertion and Reason are incorrect.
So Correct option is (4)

AIIMS - 2017

PHXI06:WORK ENERGY AND POWER

355621 Two blocks of mass 3 $k g$ and 6 $k g$ respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant $k = 200 N / m$ . Initially the spring is unstretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be :-
supporting img

1 30

c m

2 25

c m

3 20

c m

4 15

c m

Explanation:

We know that
$K . E_{s y s} = K . E_{o f C . M} + K . E_{w . r . t o C . M .}$
$K . E_{s y s} = \frac{1}{2} \times 3 \times 1^{2} + \frac{1}{2} \times 6 \times 2^{2} = 13.5 J$
$V_{C M} = \frac{6 \times 2 - 3 \times 1}{9} = 1 m / s$
$K . E_{o f C M} = \frac{1}{2} \times 9 \times 1^{2} = 4.5 J$
$\Rightarrow K . E_{w . r . t o C . M} = (13.5 - 4.5) J = 9 J$
In the absence of external force on a system
$K . E_{o f C M}$ - always remains constant
So $K . E_{w . r . t o c M}$ will be converted into $P . E$
$K . E_{w . r . t o C . M} = 9 J = \frac{1}{2} K x_{max}^{2}$
$x_{max} = 0.3 m = 30 c m$

PHXI06:WORK ENERGY AND POWER

355622 Two springs have their force constants $k_{1}$ and $k_{2} (K_{2} > K_{1})$ . When they are stretched by the same force, work done is

1 Same in both the springs

2 More in spring

k_{1}

3 More in spring

k_{2}

4 Data incorrect

Explanation:

$W = \frac{1}{2} k x^{2} = \frac{F^{2}}{2 k} \Rightarrow W α \frac{1}{k} (F =$ constant $)$

PHXI06:WORK ENERGY AND POWER

355619 A spring of spring constant $2 N c m^{- 1}$ has an extension of 5 $c m$ . The minimum work done in joule in increasing the extension from 5 $c m$ to 15 $c m$ is

1 2

J

2 16

J

3 32

J

4 4

J

Explanation:

Here, $k = 2 N c m^{- 1} = 2 \times 10^{2} N m^{- 1}$
$x_{1} = 5 c m = 5 \times 10^{- 2} m$
$x_{2} = 15 c m = 15 \times 10^{- 2} m$
Work done $=$ Increase in PE of spring
$= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 2 \times 10^{2} [{(15 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = 10^{2} \times 200 \times 10^{- 4} = 2 J$
$= 10^{2} \times 200 \times 10^{- 4} = 2 J$

PHXI06:WORK ENERGY AND POWER

355620 Assertion :
A spring of force constant $k$ is cut into two pieces having lengths in the ratio $1 : 2$ . The force constant of series combination of the two parts is $3 k / 2$ .
Reason :
The spring connected in series are represented by $k = k_{1} + k_{2}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

As we know, $k = \frac{F}{l} \Rightarrow k \propto \frac{1}{l}$
$\Rightarrow \frac{k_{2}}{k_{1}} = \frac{l_{1}}{l_{2}} = \frac{1}{2}$
$k_{1} = 2 k, k_{2} = k$
In series, $\frac{1}{k^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} = \frac{1}{2 k} + \frac{1}{k} = \frac{3}{2 k}$
$k^{'} = \frac{2 k}{3}$
So, both Assertion and Reason are incorrect.
So Correct option is (4)

AIIMS - 2017

PHXI06:WORK ENERGY AND POWER

355621 Two blocks of mass 3 $k g$ and 6 $k g$ respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant $k = 200 N / m$ . Initially the spring is unstretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be :-
supporting img

1 30

c m

2 25

c m

3 20

c m

4 15

c m

Explanation:

We know that
$K . E_{s y s} = K . E_{o f C . M} + K . E_{w . r . t o C . M .}$
$K . E_{s y s} = \frac{1}{2} \times 3 \times 1^{2} + \frac{1}{2} \times 6 \times 2^{2} = 13.5 J$
$V_{C M} = \frac{6 \times 2 - 3 \times 1}{9} = 1 m / s$
$K . E_{o f C M} = \frac{1}{2} \times 9 \times 1^{2} = 4.5 J$
$\Rightarrow K . E_{w . r . t o C . M} = (13.5 - 4.5) J = 9 J$
In the absence of external force on a system
$K . E_{o f C M}$ - always remains constant
So $K . E_{w . r . t o c M}$ will be converted into $P . E$
$K . E_{w . r . t o C . M} = 9 J = \frac{1}{2} K x_{max}^{2}$
$x_{max} = 0.3 m = 30 c m$

PHXI06:WORK ENERGY AND POWER

355622 Two springs have their force constants $k_{1}$ and $k_{2} (K_{2} > K_{1})$ . When they are stretched by the same force, work done is

1 Same in both the springs

2 More in spring

k_{1}

3 More in spring

k_{2}

4 Data incorrect

Explanation:

$W = \frac{1}{2} k x^{2} = \frac{F^{2}}{2 k} \Rightarrow W α \frac{1}{k} (F =$ constant $)$

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