IIT JEE Main 2016 Physics Question No 13

IIT JEE Main 2016 Physics Question No 13

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. [Take g = 9.8 ms–2].
(1) 9.89 × 10–3 kg
(2) 12.89 × 10–3 kg
(3) 2.45 × 10–3 kg
(4) 6.45 × 10–3 kg

Solution

Total potential energy, U is given by,
U = mgh × 1000
=> U = 10 × 9.8 × 1 × 1000
Since 20% efficiency rate, hence
3.8 × 10000000 × 20/100 m = 9.8 X 10000
=> m = (9.8 × 10000)/(3.8 × 1000000 × 2)
=> m = 12.89/1000 kg
=> 12.89 × 10–3 kg

Answer

Correct answer to given problem No 13 of IIT JEE Main Physics paper of 2016 is 12.89 × 10–3 kg i.e. option number 2.

IIT JEE Main 2016 Physics Question No 12

IIT JEE Main 2016 Physics Question No 12

‘n’ moles of an ideal gas undergoes a process A->B as shown in the figure. The maximum temperature of the gas during the process will be:

IIT JEE Main 2016 Physics Question No 12

(1) 9P0V0/(2nR)
(2) 9P0V0/(nR)
(3) 9P0V0/(4nR)
(4) 3P0V0/(2nR)

Solution

The equation of the process is y = mx + c
P = -(P0/V0)V + 3P0
Now PV = nRT
therefore P = nRT/V
=> (nRT/V)=-(P0/V0)V + 3P0
=> nRT=-(P0/V0)V2 + 3P0V
=> T=-(P0/nRV0)V2 + (3P0/nR)V                  … (1)
Now for maximum T,
dT/dV = 0                                                       … (2)
here dT/dV = -(2P0/nRV0)V + (3P0/nR)       … (3)
from (2) & (3) we get,
V=(3/2)V0
putting in (1) we get
T=-(P0/nRV0)(9/4V02) + (3P0/nR)(3/2)V0
=> T=(9/2)(P0V0/nR)-(9/4)(P0V0/nR)
=> Tmax = (9/4)(P0V0/nR)

Answer

Correct answer to given problem No 12 of IIT JEE Main Physics paper of 2016 is Tmax = (9/4)(P0V0/nR) i.e. option number 3.

IIT JEE Main 2016 Physics Question No 6

IIT JEE Main 2016 Physics Question No 6

Choose the correct statement:
(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio single.
(2) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
(3) In amplitude modulation the amplitude of the high ‘frequency carrier wave is made to very in proportion to the amplitude of the audio signal.
(4) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

Solution

By definition of Amplitude Modulation we know that statement number 3 is correct.

Answer

Correct answer to given problem No 6 of IIT JEE Main Physics paper of 2016 is option number 3 which is true.

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IIT JEE Main 2016 Physics Question No 5

IIT JEE Main 2016 Physics Question No 5

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure. Which of the following statements is false for the angular momentum L about the  origin?

IIT JEE Main 2016 Physics Question No 5
(1) L =  mv[(2/√2) + a] kˆ when the particle is moving from B to C
(2) L =  (mv/√2)Rkˆ  when the particle is moving form D to A.
(3) L =  (mv/√2)Rkˆ  when the particle is moving from A to B.
(4) L =  mv[(R/√2) – a] kˆ when the particle is moving from C to D.

Solution

As L = r  X p
Hence option 2&4 are incorrect.

 

Answer

Correct answer to given problem No 5 of IIT JEE Main Physics paper of 2016 are options (2, 4), which are false.

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IIT JEE Main 2016 Physics Question No 4

IIT JEE Main 2016 Physics Question No 4

If a, b, c, d are inputs to a gate and x is its output, then, as per the  following time graph, the gate is :

IIT JEE Main 2016 Physics Question No 4

(1) OR
(2) NAND
(3) NOT
(4) AND

Solution

As is seen from the timing graph, which is identical to that of an OR gate. In OR gate, if all the inputs are low (zero), only. Then only, the output is low (zero).

Hence, the timing graph shown is that of OR Gate.

Answer

Correct answer to given problem No 4 of IIT JEE Main Physics paper of 2016 is option (1).

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IIT JEE Main 2016 Physics Question No 3

IIT JEE Main 2016 Physics Question No 3

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95 s and 92 s. If The minimum division in the measuring clock is 1s, then the reported mean time should be :

(1) 92 +/- 1.8 s
(2) 92 +/- 3 s
(3) 92 +/- 2 s
(4) 92 +/- 5.0 s

Solution

Mean value of observations i.e. time is obtained by,
Tmean = (90 + 91 + 95  + 92)/4
=> Tmean = 92s
Also absolute errors are as follows:
E1 = |92 – 90| = 2
E2 = |92 – 91| = 1
E3 = |92 – 95| = 3
E4 = |92 – 92| = 0
Therefore,  mean absolute error = (2 + 1 + 3 + 0)/4
= 1.5 s,  but since the least count of the clock is 1 second,  so it can’t count 1.5 s. Hence, it will count 2 seconds,  as an error.
(92 +/- 2) seconds is the correct option.

Correct answer to given problem No 3 of IIT JEE Main Physics paper of 2016 is option (3).

IIT JEE Main 2016 Physics Question No 2

IIT JEE Main 2016 Physics Question No 2

For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β is:
(1) α = β/(1 + β)
(2) α = β2/(1+ β2)
(3) 1/α = (1/β) + 1
(4) α = β/(1 – β)

Solution

As we know that, β = α/(1 – α)
Hence, α = β/(1 + β)

Correct answer to given problem No 2 of IIT JEE Main Physics paper of 2016 is option (1).

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