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Since this is an MCQ, lets first do a quick sanity check. The cube is floating in equilibrium in mercury so the density of mercury should be heaver than the cube itself i.e the answer is either C or D.
thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?Since this is an MCQ, lets first do a quick sanity check. The cube is floating in equilibrium in mercury so the density of mercury should be heaver than the cube itself i.e the answer is either C or D.
Now, the cube floats 42% above the surface and therefore the immersed part i.e. the part that displaces mercury is 100-42% = 58% or 0.58.
Now we can have p1/v1 = p2/v2 where
p1 is the density of mercury,
p2 is the density of iron,
and therefore p1 = p2(v1/v2)
This gives us p1 = 7900/0.58 = 13,621 kg/m^3
Since this is a MCQ question, I skipped a lot of steps and used just ratios and the ratio based on simply density=mass/vol. This is where p1v1=p2v2 comes from.thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?
thankyou so muchSince this is a MCQ question, I skipped a lot of steps and used just ratios and the ratio based on simply density=mass/vol. This is where p1v1=p2v2 comes from.
To elaborate the steps, lets first consider the immersed part where weight of solid equals the weight of liquid displaced.
[imath]W_{s}=W_{ld}[/imath]
And since w=mg, we get
[imath]m_{s}g=m_{ld}*g[/imath]\
And since density=mass/vol, we get
[imath]P_{s}V_{s}=P_{ld}V_{ld}[/imath] as g cancels out
Now volume of liquid displaced is equal to the volume of the immersed portion, so lets write it as
[imath]P_{s}V_{s}=P_{ld}V_{imm}[/imath]
Now we rearrange as
[imath]\frac{P_{s}}{P_{l}}=\frac{V_{imm}}{V_{s}}=0.58[/imath]
And so
[imath]P_{l}=\frac{P_{s}}{0.58}=\frac{7900}{0.58}=13621\:kg\:m\:^{-3}[/imath]
Hope that helps!
Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
wont we subtract 0.94 from 1D as it says DECREASED BY 94%Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
And cross sectional area [imath]A = \frac{πd^{2}}{4}[/imath]
When the wire is stretched, the diameter is reduced to 94%.
This causes the area of stretched wire to be
[imath]A_{s} = \frac{π(0.94d)^{2}}{4} = 0.8836(\frac{πd^{2}}{4})[/imath] = 0.8836A
Volume of the wire is cross-sectional area x length,
[imath]V = AL[/imath]
And since the volume of the wire is unchanged,
[imath]AL=A_{s}L_{s}[/imath]
and the diameter is reduced to 94%, the length has to be
[imath](\frac{πd^{2}}{4})L=(\frac{π(0.94d^{2})}{4})(L_{s})[/imath]
[imath]L_{s}=\frac{L}{0.94^{2}}[/imath]
So, the length of the stretched wire is [imath]\frac{L}{0.94^{2}}[/imath] or [imath]1.132L[/imath]
Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
So Resistance of stretched wire
[imath]R_{s} = \frac{ρL_{s}}{A_{s}} = \frac{ρ(1.132L)}{0.8836A} = 1.28\frac{ρL}{A} = 1.28R[/imath]
Hope this explains!
It says decreases to 94%!wont
wont we subtract 0.94 from 1D as it says DECREASED BY 94%
You need to take intersections of both waves where they are going in the same direction.View attachment 70518
CAN SOMEONE EXPLAIN THIS QUESTION?
During one oscillation, point A moves a distance of 80mm as given in the question.View attachment 70537
AND THIS QUESTION AS WELL THANKYOU. HOW WILL WE FIND THE AMPLITUDE IN THIS QUESTION?
You need to take intersections of both waves where they are going in the same direction.
From the graph, these can be either 0 to 0.5 or it can be 0.5 to 0.8.
We can then use ratios to calculate the phase difference
[imath]\frac{\Delta t}{T} = \frac{\Delta \phi}{360^{\circ}}[/imath]
Since T is 0.8 i.e one full cycle, using [imath]\Delta t = 0.3[/imath] gives us [imath]\Delta \phi = 135^{\circ}[/imath] and using [imath]\Delta t = 0.5[/imath] gives us [imath]\Delta \phi = 225^{\circ}[/imath]
So the answer is C i.e 135 which is also 360-225 i.e. 13
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