# (i)A glass flask having mass 309 g and as interior volume of 500 cm^3 floats on water,when it is less than half filled with water.the density of material of flask is (ii)A open U tube contains Hg When 11.2 cm of water is poured into one of ams of the tube,how high does the Hg rise in other arm from its initial level (iii)A beaker of water is palced on platform of spring balance .The balance reads 1.5 kg.a stone of mass 0.5 kg and d 500kg./m^3 is immersed in water without touching the walls of the beaker.What will the balance read now

1
by rajusetu
PLZ SOLVE it
please do not write somany queries in one question. if it is a question of points, we see later ok.
it is difficult to moderate.
sir i will club questions but increase numbere of points.Is it okay?

2014-12-26T00:47:51+05:30

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Mass m of empty flask = 309 gms
Internal volume (free space) = 500 cc  cubic cm
Let total external volume of flask be = V

As the flask half filled (250 cc) with water just gets immersed fully in water, it means that the weight of water displaced is equal to the total weight of flask + filled water.

V * 1 gm/cc * g = (309 gm + 250 cc * 1 gm/cc) * g
V = 559 cc

volume of material in flask = external volume - internal empty space

density of material of flask =  mass / volume = 309 / (559 - 500)  gm/cc
= 309 / 59  gm / cc
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Initially in both arms let the height be = H cm from the bottom of U tube. So there is 2*H cm length of mercury in the tube. Area of tube is A and acceleration due to gravity is g.

Let water be poured in one arm measuring 11.2 cm in height. Due to this  the mercury in that arm goes down by h cm compared to previous position.  The mercury in the other arm goes up by h cm, compared to previous position.

pressure at the bottom of each arm is same, as per Pascal's law.

(H+ h) * 13.6 gm/cc * A * g
=  (H - h) * 13.6 gm/cc * A * g + 11.2 cm * 1 gm/cc * A * g
Solving  2 * h * 13.6 = 11.2
h =  5.6/13.6  cm
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Let us say that the solid (it is lighter than water as its density is less than 1000 kg/m^3) is immersed in water.  Asthe solid is lighter, it does not stay inside water. It will try floating up.  It is not clear it is being held immersed inside or not.

What ever the case is, the weight shown by the spring balance will increase.  If the solid is left with out holding.  It will float on water.  Then,  the weight of solid is the additional weight to the system.  So the balance will read 1.5 + 0.5 = 2 kg.  The forces of buoyancy are internal to the water and solid.  For external system as balance, only the gravitational forces on each will count.

Solid will replace 0.5/500 * 10^6 cc = 1000 cc of water.  So water level will go up by 1000 / A, where A is the area of  of cross section of beaker.  The additional pressure due to that on the bottom of beaker will be = 1000/A * 1 gm/cc * g.  Additional force on the bottom of beaker by water will be  Pressure  * area =  1000 *g in CGS units or 1 kg weight.  Hence, the balance will show  1.5 kg + 1 kg = 2.5 kg.

Additional weight from  2kg to 2.5 kg -- can be explained because there is some force pushing the solid down  to remain inside water.  That is equal to 0.5 kg force.