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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.

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

thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?

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

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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?

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!

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!

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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!

wont we subtract 0.94 from 1D as it says DECREASED BY 94%

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!

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It sayswont

wont we subtract 0.94 from 1D as it says DECREASED BY 94%

If it were

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You need to take intersections of both waves where they are going in the same direction.View attachment 70518

CAN SOMEONE EXPLAIN THIS QUESTION?

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. 135

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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?

And during this one oscillation, point A would return to its original position. Any point would move a distance = 4 time the amplitude during one oscillation, .

Remember, as defined above, the displacement is up and down while the wave travels to the right (in this case). Amplitude is the maximum displacement form the equilibrium position.

Consider the point A for example, which is originally on the minima. During one oscillation, it will move

1. a distance

2. another distance

3. another distance

4. and finally, another distance

Thus, amplitude = 80/4 = 20mm

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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PlanetMaster can u pls explain this uestion? thankyou!

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Hey redox233, I am not sure if you still need help on this, but let me explain this to you:-PlanetMaster can u pls explain this uestion? thankyou!

[imath]\rightarrow[/imath] This is a classic problem of standing waves.

[imath]\rightarrow[/imath] Given that the lowest sound produced is [imath]92[/imath] Hz, such as a node at the mouthpiece and an antinode at the other open end.

Hence, this is what we have,

We know that distance between a node and an antinode is [imath]\lambda/4[/imath].

[imath]\therefore \lambda/4[/imath] corresponds to [imath]92 \space \mathrm{Hz}[/imath].

The next arrangement of the standing wave to produce another frequency is as follows,

Here, we have [imath]\lambda/2 + \lambda/4 = 3\lambda/4[/imath]

[imath]\therefore 3\lambda/4[/imath] corresponds to [imath]3 \times \lambda/4 = 3 \times 92 =[/imath] [imath]276 \space \mathrm{Hz}[/imath]

Here is the next arrangement and you can follow this idea to get the overall correct answer to this question,

Here, we have [imath]\lambda + \lambda/4 = 5\lambda/4[/imath]

[imath]\therefore 5\lambda/4[/imath] corresponds to [imath]5 \times \lambda/4 = 5 \times 92 =[/imath] [imath]460 \space \mathrm{Hz}[/imath]

[imath]\rarr[/imath] If you simply look at all the options now in the question, we can clearly see that

[imath]\rarr[/imath] However, feel free to visualize the next arrangement and its corresponding frequency and match it with option

I hope this helps.

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No worries pal. For both waves, we can take the points they intersect 0 displacement in the same direction.

Like this:View attachment 70545

Hey PlanetMaster,

What software do you use to create such graphs that go with the background of the XPC theme? 🤔

Also, it would be great if you could provide a Math editor in the options so that we don't have to externally use LaTeX to write Mathematical equations here.

Thank you! ^_^