##### Questions: 01 to 10

**1. Starting from the same initial conditions, an ideal gas expands from volume V**_{1 }to V_{2} in three different ways, the work done by the gas is W_{1} if the process is purely isothermal, W_{2} if purely isobaric and W_{3} if purely adiabatic, then

(A) W_{2}>W_{1}>W_{3}

(B) W_{2}>W_{3}>W_{1}

(C) W_{1}>W_{2}>W_{3}

(D) W_{1}>W_{3}>W_{2 }

Answer

Option (A) is correct

**Area under the curve in a p-V diagram is the Work done.**

Considering the figure shown below, Area under the curve 2 is more than 1 and 3

A_{2}>A_{1}>A_{3}

where,

A_{1}= Area under curve 1

A_{2}= Area under curve 2

A_{3}= Area under curve 3

**From the Statement above, We have**

W_{2}>W_{1}>W_{3}

**2. A beaker filled with hot water in a room cools from 70°C to 65°C in t**_{1} minutes, 65°C to 60°C in t_{2} minutes and from 60°C to 55°C in t_{3 }minutes. Then

(A) t_{1}>t_{2}>t_{3}

(B) t_{1}=t_{2}=t_{3}

(C) t_{1}<t_{2}<t_{3}

(D) cannot be concluded

Answer

Option (C) is correct

**Considering common atmospheric condition. As temperature difference is more, heat transfer rate is higher. Hence, less time is required to cool the beaker at high temperature when compared to low temperature.**

**Note:Temperature difference is taken with respect to atmospheric temperature, so the temperature difference is not 5 but temperature drop is 5ºC.**

**If atmospheric temperature is 20ºC then initial temperature difference is 70-20=50ºC but when temperature is 65ºC than difference is 65-20=45ºC.**

**3. A house refrigerator with its door open is switched on in a closed room. The air in the room is**

(A) cooled

(B) remains at same temperature

(C) heated

(D) heated or cooled depending on atmospheric pressure

Answer

Option (C) is correct

**Generating net cold is impossible – It’s a general thermodynamic fact. So just like any other machine, refrigerators generate heat. When you first open the door you’ll get a burst of cold air, but that’s all. It’ll cool the room a little, but also heat it up a lot more. The net result would be an increase in room temperature due to the constantly running compressor.**

As the door is open the heat Q_{2} is absorbed __from the entire room__. While the heat Q_{1} is rejected to the room from the condensor.

Now we have already know that:

Q_{1} > Q_{2}

**That is, the heat rejected is more than heat absorbed in the room.**

**Hence there is a **__net heat addition__** in the room and the **__temperature of room would increase__**.**

**4. An elevator has a mass of 5000 kg. When the tension in the supporting cable is 60 kN, the acceleration of the elevator is nearly**

(A) 8 m/s^{2}

(B) 12 m/s^{2}

(C) -2 m/s^{2}

(D) 2 m/s^{2}

Answer

Option (D) is correct

Net force = 60-50=10kN

**Equating the forces, we have**

∴\(F=ma=10 \times 10^3=10^4 \;N\)

\(\large a=\frac{10^4}{5000}=2 ms^{-2}\)

**5. The piston of a steam engine moves with simple harmonic motion. The speed of rotation of crank is 120 rpm with a stroke of 2 m. What is the velocity of piston when it is 0.5 m from the centre?**

(A) 4π√3

(B) π√3

(C) 2π√3

(D) 3π√3

Answer

Option (C) is correct

**Given Data:**

*N=120 rpm; L=2m; x=1.5m; r=L/2=1m; *

**Displacement of piston from top dead center is given by**

*\(x=r(1-\cos\theta)\)*

*\(1.5=1(1-\cos\theta)\)*

*\(\theta=120^{\circ}\)*

**The velocity of piston is given by**

\(\large V=r\omega\left ( \sin\theta+\frac{\sin2\theta}{2n} \right )\)

Taking an approximation, we have

\(V=rw\sin\theta \)

\(\large V=1\times \frac{2\pi N}{60}\sin120^{\circ} \)

\(\large V=\frac{2\pi \times 120}{60}\times (\sqrt 3/2)\)

\(V=2\pi \sqrt 3 \;m/s\)

**6. The sine of the angle between the two vectors ***a = 3i + j + k *and *b = 2i – 2j *+ *k *is

(A) \(\large \sqrt{\frac{74}{99}}\)

(B) \(\large \sqrt{\frac{25}{99}}\)

(C) \(\large \sqrt{\frac{37}{99}}\)

(D) \(\large \sqrt{\frac{5}{51}}\)

Answer

Option (A) is correct

\(a=3i+j+k; \;b=2i-2j+k \)

**Dot product of two vectors is given by**

\(\large cos\theta=\frac {a.b}{|a||b|} \)

\( \large =\frac {5}{3\sqrt {11}}=\frac {5}{\sqrt {99}}\)

**We know that**

\( sin^2\theta + cos^2\theta =1 \)

\( \large sin\theta = \sqrt{1-cos^2\theta}\)

\( \large sin\theta = \sqrt {1-\frac {25}{99}}=\sqrt {\frac{74}{99}}\)

**7. Equation of the line normal to function \(\large f(x)=(x-8)^{2/3}+1\) at ***P(0,5) *is

(A) *y=3x-5*

(B) *3y=x-15*

(C) *3y=x+15*

(D) *y=3x+5*

Answer

Option (D) is correct

\(\large f(x)=y=(x-8)^{\frac {2}{3}}+1 \)

**Slope of above line is given by**

\(\large \;m_1 =\frac {dy}{dx}=\frac{2}{3}(x-8)^{\frac {-1}{3}} \)

at (0,5) \(m_1=-1/3\)

**We know that for two perpendicular lines, we have**

\( m_1 \times m_2 = -1 \)

\( m_1=-1/3 ; \; m_2=3 \)

\( y=mx+c \)

\( y=3x+c \; \)

at (0,5), we have

\( 5=c \)

\(y=3x+5\)

**8. There are 20 locks and 20 matching keys. Maximum number of trials required to match all the locks is**

(A) 190

(B) 210

(C) 400

(D) 40

Answer

Option (A) is correct

**There are 20 locks and 20 matching keys. A maximum of 19 Keys needs to be checked to find the right one for the First Lock. 18 Keys needs to be checked to find the right one for the Second Lock. This goes on, forming an arithmetic Progression with n=19**

That is: 19,18,17,…,1,0

Let a_{n}=19,18,…,1,0

**Sum of the above AP is the required answer**

\(\large S_n=\frac {n(n+1)}{2}=\frac {19(19+1)}{2}=190\)

**The official answer even after revision by ISRO authorities was B i.e. 210, they have started with the first term in AP as 20**

That is: 20,19,18,17,…,1

Let a_{n}=20,19,18,…,1

**Sum of the above AP is the required answer**

\(\large S_n=\frac {n(n+1)}{2}=\frac {20(20+1)}{2}=210\)

**9. If ø(x,y,z) is a scalar function and if ∇**^{2}f=0, then ø is

(A) Irrational

(B) Harmonic

(C) lrrotational

(D) Solenoidal

Answer

Option (B) is correct

If ø(x,y,z) is scalar function and \(\large \frac{\partial^2 \phi }{\partial x^2}+\frac{\partial^2 \phi }{\partial y^2}+\frac{\partial^2 \phi }{\partial z^2}=0\) ,then ø is harmonic.

**10. A and B are two candidates appearing for an interview by a company. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. The probability of B getting selected is**

(A) 0.9

(B) ≤0.3

(C) ≤0.6

(D) 0.5

Answer

Option (C) is correct

p(A)=0.5; p(A∩B)=0.3

**We know that**

p(A∩B)=p(A).p(B)

0.3=p(A).p(B)

0.3=0.5 x p(B)

p(B)=0.6

**Probability that both A and B are selected is at most 0.3**

So, p(B) ≤0.6

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