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1 Alternating Currents

A coil of inductance�5.0�mH�and negligible resistance is connected to an alternating voltage�V=10sin(100t). The peak current in the circuit will be:

1) 2�amp
2) 1�amp
3)�10�amp
4)�20�amp

�

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Inductive reactance $${ X }_{ L }=\omega L\\ =100\times 5\times { 10 }^{ -3 }=0.5\Omega $$

�

$${ V }_{ max }=Peak \ Voltage=10V$$

�

Peak Current $${ I }_{ p }=\dfrac { PeakVoltage }{ { X }_{ L } } =\dfrac { 10 }{ 0.5 } =20Amps$$

�

2 Surface Chemistry

If the half-life of an isotope X is 10 yr, its decay constant is

1) 6.932 yr ^-1
2) 0.6932 yr ^-1
3) 0.06932 yr ^-1
4) 0.006932 yr ^-1

Option 1

Option 2

Option 3 Solution:

SOLUTIONS

Option 4

3 Stoichiometry

1 mol of KClO₃ is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many moles of Al₂O₃ are formed?

1) 1
2) 2
3) 1.5
4) 3

Option 1 Solution:

Solution

Thermal decomposition of KClO₃
2KClO₃ → 2KCl +3O₂
Thus Two moles of KClO₃ produces Three moles of oxygen
1 mole of KClO₃ will produce 1.5moleof Oxygen
Now when Al burn 4Al +3O₂ → 2Al2O₃
Al is in excess thus oxygen is limiting reagent
Now 3 moles of Oxygen produce 2 moles of Al₂O₃
× 1.5 moles of Oxygen produce 1 mole of Al₂O₃

Option 2

Option 3

Option 4

4 Probability

An aircraft has three engines A, B and C. The aircraft crashes if all three engines fail. The probabilities of failure are 0.03, 0.02 and 0.05 for engines A, B, and C respectively. What is the probability that the aircraft will not crash?

1) 0.00003
2) 0.00090
3) 0.99997
4) 0.90307

Option 1

Option 2

Option 3 Solution:

Solution

P(aircraft will crash)=0.03 × 0.02 × 0.05 = 0.00003
P(air craft will not crash)=1−0.00003 = 0.99997

Option 4

5 Solutions

To what volume, in mL, must 50.0 mL of 3.50 M H₂?SO₄? be diluted in order to make 2 M H₂?SO₄??

1) 25
2) 60.1
3) 87.5
4) 93.2

Option 1

Option 2

Option 3 Solution:

Solution

No. of moles should be same in both cases so

M1?V1?=M2?V2?

or V2 ?= M1?V1?/M2? = 3.5×50/2 = 87.5ml

Option 4

6 Sets relations and functions

If is equal to

1) (f(x))²
2) 2f(x)
3) - 2f(x)
4) 2f(x²)

Option 1

Option 2 Solution:

SOLUTION

= 2f(x)

Option 3

Option 4

If the pressure of gas is increased from 1.01�10 ⁵ pa to 1.165�10⁵ pa and volume is decreases by 10% at constant temperature then the Bulk modulus of the gas is

(1)15.5 � 10⁵ pa (2) 1.4 � 10⁵ pa

(3) 1.55 � 10⁵ pa (4) 0.0155 � 10⁵ pa

Option 1

Option 2

Option 3 Solution:

Option 4

8 Electric Charges And Fields

A charge q is placed at the mid-point of the line joining two equal charges each of Q. If the whole system is in equilibrium, then the value of q is

Option 1

Option 2

Option 3 Solution:

SOLUTION
Potential energy of the system is equal to zero when the system is in equilibrium.

Option 4

9 Differential Equations

The degree and order of the differential equation of the family of all parabolas whose axis is x-axis are respectively-

1) 2, 1
2) 1, 2
3) 3, 2
4) 2, 3

Option 1

Option 2 Solution:

SOLUTION

Equation is y² = 4a(x – h)

differentiate we get

2y dy/dx = 4a

Again differentiate then

yd²y/dx² + (dy/dx)² =0

Clearly degree 1 and order 2

Option 3

Option 4

10 Waves

An earthquake generates both transverse (S) and longitudinal (P) sound waves in the earth. The speed of S waves is about 4.5 km/s and that of P waves is about 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicentre of the earthquake is located at a distance about

1) 25 km
2) 250 km
3) 2500 km
4) 5000 km

Option 1

Option 2

Option 3 Solution:

Solution:

Velocity of the S waves is v1 = 4.5 km/s. The velocity of the P waves is v2 = 8.0 km/s.

Let the time taken by the S and P waves to reach the seismograph be t1 and t2 . It is given that

t1 - t2 = 4 min = 4 × 60 = 240 sec ... (i)

Let the distance of the epicentre (km) be S. Then

S = v1 t1 = v2t2

$$\Rightarrow 4.5\times t_{1}-8t_{2}=0 \Rightarrow t_{2}=.4.58t_{1}$$ ... (ii)

Using (i) and (ii)

$$t_{1}-t_{2}=240 \Rightarrow t_{1}\left(1-\frac{4.5}{8}\right)=240$$

$$\Rightarrow t_{1}=\frac{240\times8}{3.5}=548.5 s.$$

$$\therefore S = v_{1}t_{1} = 4.5 × 548.5 = 2468.6 \approx 2500$$ km.

Option 4

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