Tag: energy and its forms

Questions Related to energy and its forms

A body of mass $10kg$ is moving along positive $x-$axis with $5\ m/s$ at $t=0$ and is moving along negative $x-$axis with same speed at $t=10\ s$. Average power of the force acting on the body is:

  1. $Zero$

  2. $25\ W$

  3. $50\ W$

  4. $100\ W$


Correct Option: A
Explanation:

since there is no acceleration 

there is no power
P=0

A light bulb has the rating 100 W, 220 V. If the supply voltage is 110 V, then power consumed by the bulb is

  1. 50 W

  2. 75 W

  3. 25 W

  4. 20 W


Correct Option: C
Explanation:
$P =\dfrac{ V^2}R$  

If $V = 220$ V we have

$100 W =\dfrac{ 220^2}R$

$R = \dfrac{220^2}{100} Ω = 484 Ω$. This is the resistance of the bulb.

When $V = 110$ V, power consumed $=\dfrac{V^2}R= \dfrac{110^2}{484} = 25$ W.

So, 25 W power is consumed when it is operated on 110 V.


A force $\vec {F}=(3\hat {i}+4\hat {j})N$ acts on $2kg$ movable object that moves from an initial position $\vec {r} _{1}=(-3\hat {i}-2\hat {j})m$ to a final position $\vec {r} _{1}=(5\hat {i}+4\hat {j})m$ in $6s$. The average power delivered by the force during the interval of $6s$ is equal to :

  1. $8\ watt$

  2. $\dfrac{50}{6}\ watt$

  3. $15\ watt$

  4. $\dfrac{50}{3}\ watt$


Correct Option: A
Explanation:

 

Given,

Force, $\vec {F}=(3\hat {i}+4\hat {j})N$

Displacement, $\vec{d}={{\vec{r}} _{2}}-{{\vec{r}} _{1}}=(5\hat{i}+4\hat{j})-(-3\hat{i}-2\hat{j})=\left( 8\hat{i}+6\hat{j} \right)\,m$

Work, $W=\vec{F}.\vec{d}=\left( 3\hat{i}+4\hat{j} \right)\left( 8\hat{i}+6\hat{j} \right)=48\,J$

Average power $P=\dfrac{W}{t}=\dfrac{48}{6}=8\,W$

Average power is $8\,W$ 

A weight lifter lifts $300\ kg$ from the ground to a height of $2$ meter in $3$ seconds. The average power generated by him is:-

  1. $5880\ watt$

  2. $4410\ watt$

  3. $2205\ watt$

  4. $1960\ watt$


Correct Option: D

Human heart pumps $70\ cc$ of blood at each beat against a pressure of $125\ mm$ of $Hg$. If the pulse frequency is $72$ per minute, the power of the heat is nearly.

  1. $1.2\ W$

  2. $1.4\ W$

  3. $1.6\ W$

  4. $1.8\ W$


Correct Option: B

A force F acting on a body depends on its displacement $S$ as $F \propto S^{1/3}$. The power delivered by $F$ will depend on displacement as:

  1. $S^{2/3}$

  2. $S^{-5/3}$

  3. $S^{1/2}$

  4. $S^0$


Correct Option: A
Explanation:

We know Power=$\dfrac{work \ done}{time}$

Also work done = Force $\times$displacement

As given $F \ \alpha \ {S}^{1/3}$

$F=KS^{ 1/3 }$------(1) (where $K$= constant of proportionality)

Now Power= $\dfrac{F \times S}{time}$----(2)
Putting value of 1 in 2 we get 

$P= {\dfrac{KS^{\dfrac{2}{3}}}{t}}$

An engine of ine metric ton is going up an inclined plane, 1 in 2 at the rate of 36 kmph. If the coefficient of friction is $1/ \sqrt{3}$, the power of engine is 

  1. 9.8W

  2. 98W

  3. 980W

  4. 98kW


Correct Option: D

A train of mass $6 \times 10^2$ metric tones is pulled by a locomotive. The speed of the train will be $36 \,kmhr^{}-1$. The locomotive pulls the train on the train on the level track, whose mass is $125$ metric tones. The force of friction acts on the locomotive and the train is $1 \times 10^1$ newton per metric tonne. Calculate the power of the locomotive.

  1. 72500

  2. 6000

  3. 5000

  4. 4000


Correct Option: A
Explanation:

$m = 600$
$m\ell = 125$
$m \,net = 725$
$f = \dfrac{10}{tan}$
$t _{net} = 10 \times 125$

$V = 36 \,km/h = 36 \times \dfrac{5}{18} m/s$

$P = t _{net} \times V = 72500 \,W$

The power of water pump is The power of water pump is  $4kW.$  If  $\left( g=10{ m }{ { s }^{ -2 } } \right) ,$  the amount of water it can raise in $1$ minute to a height of $20 m$  is then

  1. $100$ litre

  2. $1000$ litre

  3. $1200$ litre

  4. $2000$ litre


Correct Option: C
Explanation:

Given that,

Power, $P=4\,kW=4000\,W$

Height, $h=20\,m$

Time, $t=60\,\sec $

$ power=\dfrac{work}{time} $

$ 4000=\dfrac{mgh}{t} $

$ 4000=\dfrac{m\times 10\times 20}{60} $

$ m=1200\,Kg $

The pump can raise 1200 litre in one minute

An object of mass accelerates uniformly from rest to a speed $v _f$ in time $t _f$ Then the instantaneous power delivered to the object,as a function of time $t$ is -

  1. $mt\left(\dfrac{{v _f}^2}{t _f}\right)$

  2. $mt\dfrac{v _f}{t _f}$

  3. $\dfrac{1}{2}mt^2\left(\dfrac{v _f}{t _f}\right)^2$

  4. $\dfrac{1}{2}mt^2\left(\dfrac{v _f}{t _f}\right)$


Correct Option: A