Ans.5
from the figure you can see that what is happening actually.
now volume of the aluminium block= V
part of the volume which is submerged in the water= V1
so the buoyent force on the aluminium block = V1.ρ.g
in which ρ is representing the density of the water and g is acceleration due to gravity.
now as you can see that direction of weight of the block is in downward direction and the direction of bouyent force is in upward direcion hence the there will be a net decrease in the actualt weight of the aluminium block and becasue of this there comes a apparent weight into play.
so the apparent weight of the block= actual weight of the blcok - bouyent force on the block
W' = W-V1.ρ.g
= V.ρ'.g-V1.ρ.g-------------------(2)
now in the eqution (2)
V= volume of the aluminum block
ρ'= density of the aluminium block
V1= volume of the submerged part of the aluminium block
ρ= density of the water
g= acceleration due to gravity
now all the values has been given so put the values in equation (1) and get the answer.
Ans.7
now in this question the electron drops from second energy level to the first one hence it will emit energy in form of radiation and we know that amount of enrgy emitted by an electron from n2 shell to n1 shell is 13.6(1/n1^2-1/n2^2).Let's right it in form of equation.
emitted energy(E)= 13.6(1/n1^2-1/n2^2)
and we also know that enegy of a photon having wavelength λ is hc/λ.
so the equation will become
hc/λ=13.69(1/n1^2-1/n2^2) ------------------------(2)
in equation (2)
h= plank's constant
c= velocity of light
λ= wavelength of emitted photon
n1= value of shell in which electron has been dropped.
n2= value of shell from which electron has been emitted.
Ans.8
now in the above figure
V=potential difference across the whole circuit
I1= current in resistor R1
I2= currenti in resistor R2
I3= currnt in resisror R3