To evaporate the water that oozes out, heat is required. Part of this heat is taken from the water inside which therefore become cool. Therefore (C) is correct. Here evaporation is less, so cooling too is less.

Before transferring the fish, the position is the man with the fish and the bucket of water. After transferring the position is: the man without the fish, but the fish inside the water in the bucket. Thus to know  whether there is any change in weight, it is enough to consider whether there is any change in weight when the fish in air is transferred into the water. The answer is there is no change as when in water though it loses weight due to upthrust of water on it, the level of water increases and exerts an equal force at the bottom.So, (c) is correct.

The water rise in a capillary tube dipped in water is such that the weight of water in the tube above the surface of water in the vessel is equal to (balances) the upward force due to surface tension. So it is clear that if the the bore (radius) to the tube is decreased, the height of the water column in it must be greater.  So, (a) is correct.

When the piece of iron is lowered into the water, the water level in the bucket increases. And so the pressure at the bottom of bucket also increases. Thus the total thrust at bottom also increases. Therefore the spring balance reading increases. Thus (a) is correct.

Efficiency of Inlet pipe A = 4.16%       10024 

Efficiency of Inlet pipe B = 5.83%      1001717  

Therefore, Efficiency of A and B together = 100 %   

Now, if the efficiency of outlet pipe be x% then in 10 hours the capacity of tank which will be filled = 10 *  (10 - x)

Now, since this amount of water is being emptied by 'C' at x% per hour, then

 10x10-xx=2.5 hrs 

 => x = 8 

Therefore, in 10 hours 20% tank is filled only. Hence, the remaining 80% of the  capacity will be filled by pipes A and B  in 80/10 = 8 hours 

Combined efficiency of all the three boats = 60 passenger/trip 

Now, consider option(a)

15 trips and 150 passengers means efficiency  of B1 =  10 passenger/trip 

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip  

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

    AmitBharatNo.of Days4540Efficiency2.22Percent(=145)2.5 Percent(=140)

Amit did the work in 56 days = 56×145×2=2845 

Therefore, Rest work 17/45 was done by Amit and Bharath = 174517360 = 8 days

( since Amit and Bharath  do the work in one day = 145+140=17360)

Let the length of the train be L metres and speeds of the train Arun and Sriram be R, A and S respectively, then 

       LR-A=36 ---------- (i) 

and LR+K=24 ---------(ii)

From eq.(i) and (ii)

              3(R - A ) = 2 (R + K) 

⇒  R = 3A + 2K 

In 30 minutes (i.e 1800 seconds), the train covers 1800R (distance) but the Arun also covers 1800 A (distance) in the same time. Therefore distance between Arun and Sriram, when the train has just crossed Sriram

               = 1800 ( R - A) - 24 ( A + K)

 Time required = 1800(R-A)-24(A+K)(A+K)

= (3600 - 24) = 3576 s