Physics Questions and Answers

f= -15cm (diverging mirror);  u=35cm; v=?

⇒\(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

⇒ \(\frac{1}{-15}\) = \(\frac{1}{35}\) + \(\frac{1}{v}\)

⇒ \(\frac{1}{15}\) - \(\frac{1}{35}\) = \(\frac{1}{v}\)

⇒ \(\frac{-2}{21}\) = \(\frac{1}{v}\)

 = \(\frac{-21}{2}\) = -10.5cm

∴The image is 10.5cm behind the mirror

T=800N; I=50cm=0.5m,

m=10g=0.01kg

fundamental freq: \(f_o\) =?

\(f_o\) = \(\frac{1}{21}√{T}{μ}\)

μ =\(\frac{m}{1}\)=\(\frac{0.01}{0.5}\)

⇒ \(f_o\) =\(\frac{1}{2×0.5}\)√\(\frac{800}{0.02}\)

                  \(f_o\)  ⇒√ 40,000

⇒1st overtone = 2\(f_o\) =2×200 = 400Hz

⇒2nd overtone =3\(f_o\) =3×200=600Hz

∴3rd over tone= 4\(f_o\)  =4×200=800Hz

 

For the combination in series;

⇒R1 = 35kΩ + 40kΩ = 75kΩ

R is combined with 75kΩ in parallel to give 25kΩ

=  \(\frac{1}{R_eq}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\) 

=   \(\frac{1}{25}\)     = \(\frac{1}{R}\) + \(\frac{1}{75}\) 

=  \(\frac{1}{25}\)     - \(\frac{1}{75}\) + \(\frac{1}{R}\) 

=   \(\frac{3-1}{75}\) = \(\frac{1}{R}\) 

=   \(\frac{2}{75}\)   = \(\frac{1}{R}\) 

=   \(\frac{75}{2}\)  = R

;      R = 37.5k Ω

 W = mg = 0.5 x 10 = 5 N

 Since it's light, neglect the weight of the metre rule.

 The effective tension T acting in the vertical direction = T sin 30°

 From the second condition of equilibrium, sum of clockwise moments equal sum of anticlockwise moments

 Taking moment at E 

 ⇒ T sin 30° x 30 = 5 x 50

 ⇒ ∴T=250

 T = \(\frac{250}{15}\) = 16.67N 

Ep = 2.2kV = 2200V; Es=110V; Ns=25; Np=?

=\(\frac{E_p}{E_s}\) = \(\frac{N_p}{N_s}\)
 

= \(\frac{2200}{110}\) = \(\frac{N_p}{25}\)

=110 × Np = 2200 × 25

=\(N_p\) = \(\frac{55,000}{110}\)

∴\(N_p\) = 500 turns