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