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Let the specific heat cap. of the metal = C ∴ Heat lost by the metal = heat gained by the Calorimeter + Water ∴ 0.5 x C x [100 - 25] 80 x [25 - 15] + 420 [25 - 15] 0.5 x C x 75 = 80 x 10 + 420 x 10 37.5C = 800 + 4200 C = 5000/37.5 = 133.3Jkg-1K-1

The airtight stopper prevents heat loss by convection because it is tightly sealed.

Power input = 3KW = 3000W Voltage = 1.5KV = 1500V Current (I) = P/V = 3000/1500 = 2Amps. ∴ Power loss in the cables = I2R = 2 x 2 x 20 = 80W

1/64 is the usual fraction left of any radio active material after 4 half-lives, since 1/64 = (1/2)6 This implies that 6 half lives have elapsed. Thus 6 half lives = 120 years ∴ 1 half life = 120/6 = 20 years

When a radioactive source emits radiations that are deflected by both magnetic and electric fields, the radiations are likely to be beta particles (\beta particles). Beta particles(\beta): These are high-energy, high-speed electrons (\beta-) or positrons (\beta+). They are negatively or positively charged, respectively, and are deflected by both electric and magnetic fields due to their charge and lower mass compared to alpha particles. Gamma rays (γ): These are high-energy electromagnetic radiation and are neutral, meaning they are not deflected by electric or magnetic fields. Alpha particles (\alpha): These are positively charged and consist of two protons and two neutrons. They are not significantly deflected by electric or magnetic fields due to their relatively large mass and low speed.

Inductive reactance is given by XL = WL = 2πfL XL = 2 x π x (50/π) x 2 XL = 200Ω

Efficiency = Output/input x 100/1 Efficiency = 3/(3+1) x 100/1 = 3/4 x 100/1 = 75%

The force on the conductor is maximum when the conductor lies perpendicular to the field during which the maximum force is given as; F = ILB = 0.8 x 2 x 0.5 = 0.8N

For a photoemitted electron, eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work function = 5ev - 3ev = 2ev ∴ eV = 2eV V = 2eV/e = 2V ∴ Stopping Potential = 2.0V

An astronomical telescope traditionally consists of two converging lenses with the objective having a longer focal length than the eyepiece. In this case, the 50 cm lens would serve as the objective, and the 5 cm lens as the eyepiece, which is a standard configuration for astronomical telescopes.

In a pure (intrinsic) semiconductor, the number of electrons in the conduction band is equal to the number of holes in the valence band. This is because when an electron moves from the valence band to the conduction band, it leaves behind a hole.

In thermonuclear reactions, Energy = ∆mC\(^2\) ∆m = Change in Mass C = Speed of light ∴ 5.02 x 10\(^{-27}\) - 5.01 x 10\(^{-27}\) x [3 x 10\(^8\)]\(^2\) 0.01 x 10\(^{-27}\) x 9 x 10\(^{16}\) E = 9.0 x 10\(^{-13}\)J

The velocity V = 10 + 2t\(^2\) But acceleration, a = dv/dt thus if V = 10 + 2t\(^2\) dv/dt = 4t And t = 5s then a = 4x5 = 20m/s\(^2\) ∴ a = 20m/s\(^2\)

For a simple pendulum, the period T = 2π \sqrt 1/g Since g, 2 and π are constant, => T ∝ √l ∴ T = K√l Let the original length = l thus for the length l, period T = 17s ∴ 17 = K√1 .................. [1] Again, for the new length [1 - 1.5] the period T = 8.5s ∴ 8.5 = k \sqrt [1-1.5] ...............[2] thus dividing eqn [1] by eqn[2] we have 17/8.5 = √(l/l-1.5) ∴2 = √(l/l-1.5) =>4 = (l/l-1.5) ∴l = 4[l-1.5] = 4l - 6.0 ∴4l - l = 6.0 3l = 6.0 l = 2.0m

For a vibrating sonometer wire, the frequency F = l/2l \sqrt(T/m) and if the length, l and mass per unit length, m, are constant, then Fo ∝ \sqrtT => FoK\sqrtT or Fo/\sqrtT = K => F1o/\sqrtF1 = F2o/\sqrtF2 ∴ F1o/\sqrtT1 = F2o/\sqrtT2 thus F2 = F3x\sqrt(2T1 __\sqrtT1 = = Fox\sqrt2 x \sqrtT1 \sqrtT1 ∴F2 = Fo x \sqrt2 = Fox\sqrt2