Class 12th

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New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm

New answer posted

7 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

7 months ago

0 Follower 21 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

7 months ago

0 Follower 10 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

7 months ago

0 Follower 25 Views

R
Raj Pandey

Contributor-Level 9

f' (x)= (x- (1+x)ln (1+x)/ (x² (1+x). Let h (x)=x- (1+x)ln (1+x).
h' (x)=-ln (1+x). h' (x)>0 for x∈ (-1,0), <0 for x (0, ).
h (0)=0, so h (x)≤0. f' (x)≤0. f is decreasing.

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

K? Cr? O? (x=+6), KMnO? (y=+7), K? FeO? (z=+6). x+y+z=19.

New answer posted

7 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

E°_cell = E° (cathode)-E° (anode) = 0.16-0.52=-0.36V.
lnK = nFE°/RT = 1*0.36/0.025 = 14.4. Ans is x10? ¹, so 144.

New answer posted

7 months ago

0 Follower 8 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

SF? is see-saw shape, from trigonal bipyramidal geometry.

 

New answer posted

7 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

Osmosis.

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