Tag: simple machine

Questions Related to simple machine

In a lifting machine, an effort of 500 N is to be moved by a distance of 20 m to raise a load of 10,000 N by a distance of 0.8 m. Determine the effort lost in friction

  1. 100 N

  2. 120 N

  3. 80 N

  4. 0 N


Correct Option: A

Efficiency of a machine can also be written as 

  1. $\dfrac{\text{velocity ratio}}{\text{mechanical advantage}} \times 100 $ %

  2. $\dfrac{\text{mechanical advantage}}{\text{velocity ratio}} \times 100 $ %

  3. $\dfrac{\text{mechanical advantage}}{\text{velocity of the gear}} \times 100 $ %

  4. None of the above


Correct Option: B
Explanation:

$effiency=\eta=\dfrac{Machanical-advantage}{velocity-ratio}*100$

Fill in the blank.
In the first order of lever, velocity ratio is usually __than 1 but could be __ or equal to 1

  1. greater, less

  2. less, more

  3. greater, more

  4. less, less


Correct Option: A

You do 1200 J of work with gears. If the gears do 1000 J of work, what is the efficiency of the gears?

  1. $93.33$ %

  2. $73.33$ %

  3. $83.33$ %

  4. $80.33$ %


Correct Option: C
Explanation:

$\eta=\dfrac{Load}{effort}*100=\dfrac{1000}{1200}*100=83.33\%$

You do 1260 J of work with a wheel and axle. If the wheel and axle does 1200 J of work, what is the efficiency of the wheel and axle?

  1. More than 95 %

  2. Equal to 95 %

  3. Less than 95 %

  4. Less than 83.33 %


Correct Option: A
Explanation:

$\eta=\dfrac{Load}{effort}*100=\dfrac{1200}{1260}*100=95.23\%$

It is easier to draw up a wooden block along an inclined plane than to haul it vertically, principally because:

  1. the friction is reduced

  2. the mass becomes smaller

  3. only a part of the weight has to be overcome

  4. $g$ becomes smaller


Correct Option: C
Explanation:

The vertical plane $W=mgh$
for inclined plane $W'=mghsin(Q)$
$sinQ<1$
$W>W'$
hence only one part of weight is to overcome

A pump ejects $12000\ kg$ of water at the speed of $4\ { m }/{ s }$ in $40\ second.$ Find the average rate at which the pump is working

  1. $2.4kw$

  2. $2.5kw$

  3. $2.3kw$

  4. $1.7kw$


Correct Option: A
Explanation:
$F=ma=12000\times \dfrac {4}{40}=1200\ N$
$A=F/m=1200/1200=1/10\ m/s^2$
$s=ut +\dfrac {1}{2}at^2=0+\dfrac {1}{2}\times \dfrac {1}{10}\times 40\times 40=80\ m$
then, $W=Fs=80\times 1200=96000\ J$
rate of working, $P=\dfrac {W}{t}$
$P=\dfrac {96000}{40}=2400W=2.4\ kW$

The mechanical advantage of inclined plane of angle of inclination $60^{\circ}$ is equal to :

  1. $\dfrac{2}{\sqrt{3}}$

  2. $cosec 30^o$

  3. both (1) and (2)

  4. none of the above


Correct Option: A
Explanation:

Mechanical advantage of an inclined plane at an angle $\theta$ is  $ \dfrac{Length }{height}$$ = \dfrac{l}{h} = \dfrac{1}{sin \theta}$


$ = cosec \theta = cosec 60^o = \dfrac{1}{sin 60^o} = \dfrac{2}{\sqrt{3}}$

Mechanical advantage is equal to 

  1. $2 \pi R \times 2 \pi r$

  2. $ 2 \pi R - 2 \pi r$

  3. $\displaystyle \frac{2\pi R}{2 \pi r}$

  4. None of these


Correct Option: C