IIT JEE , AIEEE Forum - Physics , Mechanics

joyfrancis

q) A piece of wax weighs 18.03 gm in air. A piece of metal weighs 17.03 gm in water. It is tied to wax and both together weigh 15.23 gm in water. Find specific gravity of wax .?

q) A small ball rises to surface at a constant velocity in a liquid whose density is four times greater than that of the material of the ball. The ratio of viscous force acting on the rising ball and it's weight is...?

q) A wooden block with a coin placed on it's top, floats in water as shown in figure1. The distances l and h are shown there . After sometime the coin falls into the water , what would be the changes in l and h.

q) The spring balance A reads 2kg with a block m suspended from it. A balance B reads 5kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in fig 2.In this situation

a)A will read > 2

b)B will read > 5

c)A will read<2 and B>5

d)A will read 2 , B will read 5.

jinu.coolboy

2)3:1

jinu.coolboy

plz reply .is the answer right?

joyfrancis

yup!..it's right!

ankitagg

answer to last question is (c)

akhil_o

1).913

expected mass in water=17.03+18.03=35.06 g
calc mass=15.23 so U=35.06-15.23=19.73 g
ie 19.73 cc of water is displaced by wax alone(since metal upthrust is already considered when we take weight in water)

so V=19.73 cc, m=18.03 g
density=18.03/19.73
=0.913

sboosy

akhil_o

thnx dude...happens with oral calculations :)

joyfrancis

thnx! , plz try the others too..

elastiboysai

q.4)

b and c

The extension of the spring will be lower because of the buoyant force on the bloc.

So A reads <2

now there will be a reaction force to the buoyant force , on the water by d bloc.

so B >5

jinu.coolboy

for constant velocity,net force on the body should be zero.
Therefore, Mg+F=V(4row)g
V(row)g+F=V(4row)g
F=3V(row)g
Therefore 3V(row0g /V(row)g
you will get 3:1

jinu.coolboy

Re:fluids

spideyunlimited

q) A wooden block with a coin placed on it's top, floats in water as shown in figure1. The distances l and h are shown there . After sometime the coin falls into the water , what would be the changes in l and h.

acc to diagram.. the coin must b little heavier than the buoyant force on the wooden piece... so when the coin slips into the water, the wooden piece will come fully at the surface since it is less dense than water.. and coin will sink in the water.
so L will decrease
now for object denser than water the coin(ie. it will sink)
first it was displacing volume of water which was equal to volume of water it was making the wooden piece displace (since it was on top of it and pushing it down). Now , the coin is not on the wooden block, so water level H falls as wooden block rises up :) ..
Now since coin falls in water, so it must in that way also displace some water, but here the volume of that water is less than it was earlier through the wooden block :) .
So water Level of H in a net result still falls down.

(and L also reduces)

spideyunlimited

q) The spring balance A reads 2kg with a block m suspended from it. A balance B reads 5kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in fig 2.In this situation
a)A will read > 2
b)B will read > 5
c)A will read<2 and B>5
d)A will read 2 , B will read 5.

See... my comrade hehe...
the weight reading on A will be less than 2 since net weight of A will be reduced due to buoyant force acting upward (things become lighter to carry when under water, remember? :) ).
Also, reading on B will be greater than 5 because of weight of liquid + beaker... plus some extra weight which is, the reduced weight of mass m :) .. so B weighs slightly more than 5.

SO OPTION (C)!

PS - elastiboy it is best not to choose (b )since it means no restriction on A... o wel thts the vagueness of such options but anyway i know u know the answer :)

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