Question Library

Resolution of force: Total vertical upward force: = 3√3 + 10 cos 60o = 3√3 + 5. Total downward force = 6 cos 30o = 6 x √(3/2) = 3√3 Total horizontal forces towards right; = 3N + cos 30o = 3N + 10 x √(3/2) = 3N + 5√3 Total horizontal forces towards left: = 6 cos 60o = 6 x 0.5 = 3.0N Summary: Vertical resultant forces upward: = 3√3 + 5 - 3√3 = 5N Horizontal resultant forces towards right: = 3N + 5√3 - 3√3 = 5N Horizontal resultant force towards right = 3N + 5√3 - 3N = 5√3N ∴ R2 = 52 + (5√3)2 = 25 + 25 x 3 = 100 ∴ R = √100 = 10N

The ball has its fastest speed at its equilibrium position when its displacement X = 0, which is at X. At this point, the bob has maximum velocity and maximum kinetic energy.

Being a uniform rod, the height W should be acting at the mid point 25cm mark. ∴ Taking moment about the fulcrum, F, 200 x 10 = W x 10 + 40 x 30 2000 = 10W + 1200 => 10W = 2000 - 1200 = 800 w = 800/10 = 80N

Effective component of the force along the horizontal = F cos 60o Work done by this force : 60 F cos 60 x X = 2 x 0.5 x 3 = 3J

Ratio of their linear expansion = \(\frac{∝_1}{∝_2}\)= 3:4. When heated to the same temperature range, the ratio of their increase in length \(\frac{e_1}{e_2}\) = 1:2 Since e = l \(\alpha \Delta T\) and \(\Delta T\) is same, \(\frac{e_1}{e_2} = \frac{l_1 \alpha_1}{l_2 \alpha_2}\) \(\frac{1}{2} = \left( \frac{l_1}{l_2} \right) \times \left( \frac{3}{4} \right)\) \(\frac{l_1}{l_2} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}\) So, \(\frac{l_1}{l_2}\) = 2:3.

The fundamental frequency is related to tension as F ∝ √T =>F= K√T ∴F1=F2=>250=500√T1√T2√400√T2 ∴√T2 = 500√400 = 2 x 20250 ∴ T2 = 402 = 1600N

Since the cells are connected in parallel the net e.m.f is that of one cell, E. However the effective internal resistance should be added as parallel resistance. => 1/v + 1/25 + 1/25 + 1/25 = 1.2 ∴ v = 0.83Ω ∴ Effective e.m.f and internal resistance are respectively = 1.5V and 0.83Ω

Use Fleming's Left-Hand Rule to find the direction: First Finger (Field): Point it into the paper/screen. Thumb (Force): Point it upwards. Second Finger (Current): It will naturally point to the right.

Period is the time to complete one cycle of 360o => Periodic Time T ≡ 1 cycle of 360o ∴ T/4 = 360/4 = 90o ∴ Instantaneous e.m.f E = Eo sin ωt = Eo sin 90 But sin 90 = 1: => E = Eo x 1 = Eo

The decay process illustrated above shows that the lost \beta -particle to decay to the X by increase its atomic number ( from 92 to 93) and leaving the mass no. unaffected. There after, x decayed by -\alpha emission to give Y, by decreasing the mass no. by 2 ( from 93 to 91). Thus , the emission is \beta and \alpha particles respectively

The inductive reactance is given as Xl = 2πfl = 2 x 50/π x 0.1 = 10Ω Again from ohms law, R = V/I = 75/1.5 = 50Ω ∴ Thus the inductive reactance and the resistance are respectively 10Ω, and 50Ω

From faradays law, M = itz But mass, M = volume x density and volume = cross-sectional area x thickness M = 118.8 x 10\(^{-6}\) x 9 x 10\(^3\) ∴ 118.8cm2 x 10\(^{-6}\) x 9 x 10\(^3\) = 10 x t x 3.3 x 10\(^{-7}\) ∴ (118.8 x 9) x 10\(^{-7}\) = 10 x t x 3.3 x 10\(^{-7}\) ∴ t = \(\frac{118.8 \times 9}{ 3.3}\) t = 324 seconds Convert to minutes by dividing by 60 t = \(\frac{324}{60}\) t = 5.4 mins.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option D.