Physics Test 6

In a discharge tube, after being accelerated though a high potential difference the ions in the gas strike the cathode with huge kinetic energy. This collision liberates electrons from the cathode. These free electrons can further liberate ions from gas molecules through collisions. The positive ions are attracted towards the cathode and negatively electrons move towards anode. Thus, ionization of gas results.

For photo emission the wavelength of incident ray should be less than threshold value. 

X-ray have wavelength less than ultraviolet rays.

In case the charge Q is moving freely through the conductor under the action of electric field, their kinetic energy would increase as they move because electric field would exert force on them. This force would be attractive/repulsive based on the nature of charge Q.

X decay to Y (stable)

Ratio of X and y = 1:7

at the start t₀ = 0 X = N₀ and Y = 0

after time t

X will be N₀ – X

Y = X

N₀ – X/x = 1/7

X = 7 N₀/8 

Remaining nuclei = N₀ – X = N₀ - 7 N₀/8

= N₀/8 = N₀/2³

Number of half life passed = 3

t = 3X half life = 3x 1.4 × 10⁹ years = 4.2× 10⁹ years

The elements high on the B.E. versus mass number plot are very tightly bound and hence, are stable. And the elements those are lower on this plot, are less tightly bound and hence, are unstable. 
Since helium nucleus shows a peak on this plot so, it is very stable.

P = VI 

∴ I = P/V = 360/60 = 6 A 

Now, 

V = IR 

∴R = V/I = 60/6 = 10 Ω

To work as voltmeter, the resistance of galvanometer should be high.

The voltameter is connected in parallel to the load for high efficiency minimium current should pass through it so resistance should be high ideally infinite.

For this high resistsnce wire to be connected in series with the coil.

Let R₁, R₂ and R₃ are the resistance of three bulbs respectively. 

R= R₁+R₂+R_{3 
P = V}²/R

V²/P = V²/P₁ + V²/P₂ + V²/P₃ 

1/P = 1/60 + 1/60 + 1.60

P =60/3 = 20W