Physics Questions and Answers

The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. Alpha particles can be blocked by a few pieces of paper. Beta particles pass through paper but are stopped by aluminum foil. Gamma rays are the most difficult to stop and require concrete, lead, or other heavy shielding to block them.

Therefore, X = γ-ray; Y = α-particle; Z = β-particle

The sensitivity of a thermometer is defined as the smallest temperature change that can be detected or measured, NOT how quickly that temperature change can be detected.

In contrast, the difference between the maximum and the minimum temperature is the range of a thermometer

density (ρ) = 13600 \(kgm^{-3}\); g =\(9.8 ms^{-2}\)

\(P_{atm}\) =101,325 Pa;  ρ=13600 \(kgm^{-3}\)


Pabs=\(P_{atm}\)+ρgh

⇒Pabs=101,325+13600 x 9.8 x 0.25
 

⇒Pabs=101,325+33320

⇒ Pabs=134,645Pa

\(n_1\)=1.33; \(n_2\)=1.5;  i=40°; r= ?

from snell's law:

\(n_1\) x sin i=\(n_2\)x sin r

⇒1.33 x sin 40°=1.5 x sin r

⇒sin r = 0.5700

⇒r = \(sin^{-1}\) (0.5700)

 ⇒ r = 34.75

The work done against the gravitational force is calculated using the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately \(9.8 ms^2\) on Earth), and h is the height. Substituting the given values, we get W = 3.0 kg * \(9.8 m/s^2\) * 240 m = 7.2 kJ. Therefore, the correct answer is '7.2 kJ'.

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