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Work - class-XI

Description: work
Number of Questions: 20
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Tags: work, energy and power physics work and energy energy and its forms
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The correct relation between joule and erg is:

  1. $1\ J = 10^{-5} erg$

  2. $1\ J = 10^{5} erg$

  3. $1\ J = 10^{-7} erg$

  4. $1\ J = 10^{7} erg$


Correct Option: D
Explanation:

Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.

Thus,

$1 joule = 1 newton \times 1 m\\$

$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$

$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$

$1 J = 10^{7} \ erg$

Thus option D is correct.

State the wrong statement 

  1. Total work done by internal force in a system is always zero

  2. Work done is different as seen from different frames of reference

  3. In the absence of external forces and non-conservation force , the molecules energy of a system remains conserved

  4. a non conservation force always  do negative work


Correct Option: D

A force $\vec {F} = -k(x\hat {i} + y\hat {j})$, where $k$ is positive constant, acts on a particle moving in the $x-y$ plane. Starting from the origin, the particle is taken along the positive x-axis to the point $(a, 0)$ and then parallel to the y-axis to the point $(a, a)$.

  1. Work done by the force in moving particle along x-axis is $-\dfrac {1}{2}ka^{2}$

  2. Work done by the force in moving particle along x-axis is $-ka^{2}$

  3. Work done by the force in moving particle along y-axis is $-\dfrac {1}{2}ka^{2}$

  4. Total work done by the force for overall motion is $-ka^{2}$


Correct Option: B,C

A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

  1. Movies it over an inclined plane

  2. Movies it over a horizontal surface

  3. Lift it vertically upwards

  4. All of the above


Correct Option: C
Explanation:

The maximum work done by man will be when he lift it vertically upwards because in such situation the man has to exert force opposite to gravity that is in the direction of the displacement of load. 

A force of $5 N$ is applied on a $20 kg$ mass at rest. the work done in the third second is:-

  1. $\dfrac{25}{8}J$

  2. $\dfrac{25}{4}J$

  3. $12 J$

  4. $25 J$


Correct Option: A
Explanation:

Displacement in third second = displacement till 3rd second - Displacement till 2nd second 

=$\dfrac{a}{2}3^2-\dfrac{a}{2}2^2=\dfrac{5}{8}m$ (No term of ut because u=0)

Thus work done in third second =$5\times\dfrac{5}{8}=\dfrac{25}{8}J$

A small ball bearing is releases at the top of a long vertical column of glycerine of height $2h$. The ball bearing falls through a height $h$ in a time $t _{1}$ and then the remaining height with the terminal velocity in time $t _{2}$ Let $W _{1}$ and $W _{2}$ be the work done against viscous drag over these height. therefore.

  1. $t _{1}< t _{2}$

  2. $t _{1}> t _{2}$

  3. $W _{1}=W _{2}$

  4. $W _{1}< W _{2}$


Correct Option: A

No work is done by a force on an object if 

  1. the force is always perpendicular to its velocity

  2. the force is always perpendicular to its acceleration

  3. the object is stationary but the point of application of the force moves on the object

  4. the object moves in such a way that the point of application of the force remains fixed.


Correct Option: A
Explanation:
The work done by a force F in displacing a particle through d is:
$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$
If $\theta =90^o$
$W=Fd\cos 90^o=0$
Now direction of displacement is same as the direction of velocity.
So if force is always perpendicular to the velocity then no work is done.

A moving particle is acted upon by several forces $F _1, F _2, F _3 .....$ etc. One of the force is chosen, say $F _2$,  then which of the following statement about $F _2$ will be true.

  1. Work done by $F _2$ will be negative if speed of the particle decreases

  2. Work done by $F _2$ will be positive if speed of the particle increases

  3. Work done by $F _2$ will be equal to the work done by other forces if speed of the particle does not change

  4. If $F _2$ is a conservative force, then work done by all other forces will be equal to change in potential energy due to force $F _2$ when speed remains constant.


Correct Option: A

The unit kg m$^2$s$^{-2}$ is associated with :

  1. work only

  2. kinetic energy only

  3. potential energy only

  4. all the above


Correct Option: D
Explanation:

The given unit is that of Work and Energy.


Hence all the physical quantities mentioned have the same unit.

Option D is correct
−2

The unit N-s is equivalent to:

  1. J

  2. kgms$^{-1}$

  3. kgms$^{-2}$

  4. kgms


Correct Option: B
Explanation:

unit of force in Newton($N$).

force =mass$\times$acceleration.
in units $N=kg.m.sec^{-2}$
so $N.s=kg.m.sec{-1}$
so best possible answer is option B.

Rahul took a wooden cube of volume $1000  {cm}^3$ and put it in water. He observed that $\displaystyle \frac{3}{5}th$ of its volume is below the level of water. Later, he floated the cube in a liquid of density $0.8  g  {cm}^{-3}$ and applied extra force on the cube to completely submerge it in the given liquid. Calculate how much extra force Rahul applied on the cube.

  1. 2 N

  2. 8 N

  3. 6 N

  4. 4 N


Correct Option: A

1 J is equal to

  1. ${10}^{5}$ erg

  2. ${10}^{7}$ erg

  3. ${10}^{-7}$ erg

  4. none of these


Correct Option: B
Explanation:

$1 J=1N\times1m$


$={10}^{5}dyne\times{10}^{2}cm$

$={10}^{7}erg $

The c.g.s. unit of work is

  1. N

  2. dyne

  3. erg

  4. J


Correct Option: C
Explanation:

The $c.g.s.$ unit of work is $erg.$

The $m.k.s.$ unit of work is $N$

One erg is equal to

  1. $1 g cm s^{-2}$

  2. 1 Nm

  3. $10^7 J$

  4. $10^{-7}J$


Correct Option: D

Select the correct statement for work, heat and change in internal energy

  1. Heat supplied and work done depends on initial and final states

  2. Change in internal energy depends on the initial and final states only

  3. Heat and work depend on the path between the two points

  4. All of these


Correct Option: A

How many electron volts make one Joule?

  1. $ 3.25 \times 10^{19} \mathrm{eV} $

  2. $ 6.25 \times 10^{18} \mathrm{eV} $

  3. $ 9.25 \times 10^{17} \mathrm{eV} $

  4. $ 1.25 \times 10^{20} \mathrm{eV} $


Correct Option: B

Water falls from a height of $210\, m$. Assuming whole of energy due to fall is converted into heat the rise in temperature of water would be 
($J = 4.3$ Joule/cal)

  1. $42^0\, C$

  2. $49^0\, C$

  3. $0.49^0\, C$

  4. $4.9^0\, C$


Correct Option: C
Explanation:

$\begin{array}{l} \Delta \theta =0.0023h \ =0.0023\times 210 \ ={ 0.483^{ 0 } }C \ =\approx { 0.49^{ 0 } }C \end{array}$

Hence, Option $C$ is correct.

1 J Is equal to $1 : kg : m : s^{-2}$

  1. True

  2. False

  3. Ambiguous

  4. Data insufficient


Correct Option: B
Explanation:

False
because,A Joule is equal$= Newton \times meter$, and a Newton $= mass \times acceleration$The unit for mass is kg and for acceleration is m${s}^{2}$. So if a Joule $=Newton \times meter$, it equals $kgm{s}^{2}times m$. The x meter makes the meter
on top squared so you get Joules $= kg{m^2}$${s}^{-2}$. 

Which of the following will give $1J$ of work?

  1. $F=1N,S=1m,\theta={0}^{0}$

  2. $F=1N,S=1m,\theta={90}^{0}$

  3. $F=0.1N,S=1m,\theta={0}^{0}$

  4. $F=0.1N,S=10m,\theta={90}^{0}$


Correct Option: A
Explanation:

$W=F.S.cos\theta$
For $F=1, S=1, \theta={0}^{0}$, $W=1\times1{\times}cos{0}^{0}=1 J$

$4.0\times{10}^{-19}J=$

  1. $1eV$

  2. $2eV$

  3. $2.5eV$

  4. $5eV$


Correct Option: C
Explanation:

$1eV=1.6\times{10}^{-19}J$


$\therefore{4.0}\times{10}^{-19}J=\dfrac{4.0\times{10}^{-19}}{1.6\times{10}^{-19}}eV=2.5eV$

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