Physics Questions and Answers

Let mb=mass of empty bottle,

\(m_w\)=mass of water only and

\(m_a\)= mass of alcohol only
 

given; \(m_b\)=19g



\(m_b\) + \(m_w\) = 66g

\(m_b\) + \(m_a\) = ?

R.d=0.8

R.d=mass of alcohol

⇒\(\frac{mass of alcohol}{mass of equal volume of water}\)

⇒ mass of equal volume of water = \(m_w\)=66-19=47g

⇒0.8 = \(\frac{m_a}{47}\)

⇒\(m_a\)=0.8×47 =37.6g

; \(m_b\) + \(m_a\) = 19+37.6=56.6g

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