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This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- on resolving forces into rectangular components in equilibrium forces (F₁+1/ $\sqrt[]{2}$ )N are equal to $\sqrt[]{2}N$  and F₂ is equal to ( $\sqrt[]{2}+1/\sqrt[]{2}$ )N

F₁+1/ $\sqrt[]{2}$ = $\sqrt[]{2}$

F₁= $\sqrt[]{2}-1/\sqrt[]{2}$ = $\frac{2-1}{\sqrt[]{2}}=\frac{1}{\sqrt[]{2}}=0.707N$

F₂= $\sqrt[]{2}+\frac{1}{\sqrt[]{\sqrt[]{2}}}=\frac{2+1}{\sqrt[]{2}}=\frac{3}{\sqrt[]{2}}=2.121N$

This is a assertion and reason answer type question as classified in NCERT Exemplar

At constant temperature, the pV vs V plot for real gases is not a straight line because there is intermolecular attraction present in real gases which is absent in ideal gases; hence ideal gases form a straight line in the pV vs V plot at constant temperature.

This is a assertion and reason answer type question as classified in NCERT Exemplar

option (i). Both A and R are true and R is the correct explanation of A. 

This is a matching answer type question as classified in NCERT Exemplar

(i)     -    B

(ii)    -     C

(iii)   -     A

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – horizontal velocity u_x= v_s

During projectile motion horizontal velocity remains unchanged

V_x=u_x=v_s

In vertical direction v_y²= u_y²+2gH

V_y= $\sqrt[]{2gH}$

Resultant speed of the ball at the bottom

V= $\sqrt[]{v_x^2+v_y^2}$

V= $\sqrt[]{v_s^2+2gH}$

b) when ball is given horizontal velocity and a small downward velocity

in horizontal direction V'_x=u_x=v_s

in vertical direction v_y'²= u_y²+2gH

resultant velocity of the ball at the bottom

v'= $\sqrt[]{v_s^2+u^2+2gH}$

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- displacement vector of particle is r (t)=? Acos $wt+$ ? Bsinwt

X=Acoswt

x/A= coswt………1

displacement along y axis is

y=Bsinwt

y/B= sinwt……….2

squaring and then adding eqn1 and 2 we get

x²/A²+y²/B²=cos²wt+sin²wt =1

this is an equation of ellipse. Therefore trajectory of particle is an ellipse.

b)v= dr/dt= id/dt (Acoswt)+jd/dt (Bsinwt)

 ? [A (-sinwt)w]+? [B (coswt).w]

= -? Awsinwt+? Bwcoswt

Acceleration a= dv/dt

So a= -? Awd/dt (sinwt)+? Bw [-sinwt]w

=? Aw²coswt-? Bsinwt

= -w²r

So force acting on the particle f=ma=-mrw²

The documents that are required during the Indus International University (IIU) are as follows: 

• Class 12 marksheet
• Graduation marksheet
• Class 10 marksheet 
• Government issued ID
• Transfer Certificate

Candidates with the above mentioned documents can submit the application form and successfully register for their desired programme. 

Yes, you would have to submit your GRE scores to increase chances of acceptance for the MS in Computer Science at the CWRU. It is considered beneficial if you have a UG in a computer-related field from a recognised university,and submit documents like academic transcripts, LOR, a purpose statement, and an updated CV are also crucial to increase the chances of selection.

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – centripetal force, F= mv²/r= f= $\mu N=\mu mg$

V= $\sqrt[]{\mu rg}$

For path ABC path length =3/4 (2 $\pi 2R$ )= $3\pi R=3\pi \left(100\right)=300\pi$

V₁= $\sqrt[]{\mu 2Rg}=\sqrt[]{0.1\times 100\times 10}=1.414m/s$

So t= 300 $\frac{\pi }{1.414}=66.6s$

For path DEF  path length = $\frac{1}{4}\left(2\pi R\right)=50\pi$

V₂= $\sqrt[]{\mu Rg}=\sqrt[]{0.1\times 100\times 10}=\frac{10m}{s}$

So t₂= 50 $\frac{\pi }{10}$ = 15.7s

For path CD and FA

Path length =R+R= 200m

T= 200/50= 4s

Total time = 66.6+15.7+4= 86.3s

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- v_x=2t for 0

= 2 (2-t) for 1

=0 for t>2s

V_y= t for 0

= 1 for t>1s

F_x= ma_x= mdv_x/dt= 1 (2)

F_y = ma_y= mdv_y/dt

= 1 (1) for 0

F= F_x? +F_y?

= 2? +?

=-2?

=0

The selection criteria for the courses at Indus International University (IIU) can either be merit-based or entrance-based, depending upon the programme. Candidates are required to participate in the admission process by filling out the application form on the official website of the university. 
Candidates might need to appear for entrance exams such as JEE Main or CMAT. Learn more about the selection criteria through the official website of the university. 

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- as angle is 45

On smooth inclined plane acceleration will be a = gsin $\theta$

So acceleration will become a= g/ $\sqrt[]{2}$

Using equation of motion s =ut +1/2at²

S= $\frac{1}{2}\frac{g}{\sqrt[]{2}}{T}^2$

On rough inclined plane a = g (sin $\theta -\mu cos\theta$ )

= g (sin $45-\mu cos45$ )= $\frac{g(1-\mu )}{\sqrt[]{2}}$

So s=ut +1/2at²

S= 0+ $\frac{1}{2}\frac{g\left(1-\mu \right)}{\sqrt[]{2}}\left(pT\right)$ ²

Comparing two above distance

$\frac{\mathrm{}}{}$$\frac{1}{2}\frac{g\left(1-\mu \right)}{\text{exception 25:}}\left(pT\right)$²= $\frac{1}{2}\frac{g}{\text{exception 25:}}{T}^2$

So after solving we get $$$\mu =(1-1/p^2)$

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• Practical Macro (Portfolio Management Pathway)
• Due Diligence (Private Markets Pathway)
• Managing Private Wealth Clients (Private Wealth Pathway)