VITEEE 2014 :: Physics :: General questions

• Physics - General questions

An installation consisting of an electric motor driving a water pump left 75 L of water per second to a height of 4.7 m .If the molor consumes a power of 5kW , then the efficiency of the installation is 

Answer:

Explanation:

Power consumed by motor =5kW

=  $5 \times 10^{3}W$=5000W

power used in lifting water =$\frac{mgh}{t}$

 $=7.5 \times 9.8 \times 4.7 =3454.5kW$

 Efficiency = power used/ power  consumed  x 100%

  $=\frac{3454.5}{5000} \times 100=$69%

Current in the circuit will be 

Answer:

Explanation:

Here, diode  in lower  branch is forward  and in upper branch  is reveresed  biased

$\therefore$    $i= \frac{5}{20+30}=\frac{5}{50}A$

If the nuclear fission, piece of uranium of mass 5.0 g is lost, the energy obtained in kWh is?

Answer:

Explanation:

 As we know, energy $E= \triangle mc^{2}$

$=0.5 \times 10^{-3} \times (3 \times 10^{8})^{2}$

    $=4.5 \times 10^{13}J$

  $=\frac{ 4.5 \times 10^{13}}{3.6 \times 10^{6}}=1.25 \times 10^{7} kWh$

The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V. What increase in plate potential will keep the plate current unchanged?

Answer:

Explanation:

Amplification factor  of a triode ,  $\mu=\frac{\triangle  V_{p}}{\triangle  V_{g}}$

$\Rightarrow$    $\triangle  V_{p}=-\mu\times \triangle  V_{s}=-50(-20)=10V$

• Physics - General questions