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A free body diagram needs to be drawn.

The length of the bar, l = 2 m

T1 and T2

At translational equilibrium, we have $\stackrel{?}{L_q}=\stackrel{?}{L_R}$ = $T_1$

(T1 / T2) = ( $T_2$ / $T_1\sin ?36.9°$ = 4/3

T1 = (4/3)T2

For rotational equilibrium, on taking the torque about the centre of gravity, we have

T1 $T_2\sin ?53.1°$ x d = T2 $\sin ?53.1°$  (2-d)

T1 x 0.8d = T2 x 0.6 (2-d)

(4/3)T2 x 0.8d = T2 x 0.6 (2-d)

(4/3) x 0.8d = 0.6 (2-d)

1.07d = 1.2 – 0.6d

d = 0.72

So the c.g. of the given bar lies at 0.72 m from its left end.

Let at certain instant two particles be at points P and Q, as shown in the figure.

Angular momentum of the system about point P

$\stackrel{?}{L_p}$ = mv x 0 + mv x d = mvd ……. (i)

Angular momentum of the system about point Q

$\stackrel{?}{L_q}$ = mv x d + mv x 0 = mvd ……. (ii)

Consider a point R, which is at a distance y from point Q such that QR = y

PR = d – y

Angular momentum of the system about point R

$\stackrel{?}{L_R}$  = mv x (d – y) + mv x y = mvd – mvy + mvy = mvd ……. (iii)

Comparing equations (i), (ii) and (iii) we get

$\stackrel{?}{L_p}$ = $\stackrel{?}{L_q}$ = $\stackrel{?}{L_R}$ …… (iv)

Pandit SN Shukla University has a total of 40 seats for Degree course in Electronics Science. Candidates are admitted to this programme as per the sanctioned intake and performance in the last qualifying exam. It must be noted that the mentioned seat count is as per the official website/ sanctioning body. It is subject to revision and hence, is indicative. 

2. (i) 5 $\in$ A

(ii) 8 $\notin$ A

(iii) 0 $\notin$ A

(iv) 4 $\in$ A

(v) 2 $\in$ A

(vi) 10 $\notin$ A

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Let $\stackrel{?}{c}$ = $\stackrel{?}{b}$ ,  $\stackrel{?}{a}$ = $\stackrel{?}{a}$ ,  $\stackrel{?}{OA}$ = $\stackrel{?}{b}$

Let $\stackrel{?}{OB}$ be a unit vector perpendicular to both b and c. Hence $\stackrel{?}{c}$ and a have the same direction

Now $\stackrel{?}{OC}$ = bc $\stackrel{?}{n}$ $\stackrel{?}{n}$

= bc $\stackrel{?}{b}\times \stackrel{?}{c}$ $\sin ?\theta$ = bc $\stackrel{?}{n}$

Now $\sin ?90°$ )= a. (bc $\stackrel{?}{n}$ ) = abccos $\stackrel{?}{n}$ ? = abccos0° = abc = Volume of the parallelepiped

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Let AB is equal to the vector a and AC be equal to the vector b.

Consider two vectors $\sin ?\theta$ = $\frac{CN}{AC}$ = $\frac{CN}{\left|\stackrel{?}{B}\right|}$

$\stackrel{?}{AC}$ = $\stackrel{?}{b}$ inclined at an angle $\theta$

MN = $\stackrel{?}{b}\sin ?\theta$

| $\stackrel{?}{a}$ $\times \stackrel{?}{b}$ | = | $\stackrel{?}{a}$ | | $\stackrel{?}{b}|\sin ?\theta =AB\times AC$

The area of ΔABC, we can write the relation

Area of Δ ABC = $\frac{1}{2}$ AB $\times AC$ = $\frac{1}{2}\left. [\stackrel{?}{a}\times \stackrel{?}{b}\right]$

There are over 40 seats for BSc courses offered by Pandit SN Shukla University. Candidates are admitted as per the sanctioned intake and performance in the last qualifying exam. There is no information about seat reservations. However, the state or central government rules for seat reservation may be considered. 

The child is sitting on the trolley and there is no external force, hence it is a single system. The velocity of the centre of mass will not change, irrespective of any internal motion.

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• Scholarship through MYSY scheme- Students who scored 80 percentile and above in the Class 12/10 are given scholarship in case their family income is less than INR 6 Lacs annually.
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If mass of the H atom = m, mass of the Cl atom = 35.5m

Given x + y = 1,27 À

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore,

We can have, : (my+35.5mx)/ (m+35.5m) = 0

 mx + 35.5my = 0

x = 35.5 (1.27 – x)

x = 1.24 À

So the centre of mass lies 1.24 À from H atom

All the structures specified are symmetric bodies with uniform mass density. For all these bodies, their centre of mass will lie in their geometric centres.

Not necessarily, the centre of gravity of a circular ring is at the imaginary centre of the ring.

APG Shimla BTech application process is online. Candidates have to apply online to get admission into the BTech programme. To get admission into APG Shimla BTech candidates have to apply through the university website if they are eligible to seek admission into the institution.

The document verification at APG Shimla BTech take place offline at the campus of the affiliated college. Candidates have to appear in person for the document verification and fee payment. It is only after this process admissions are finlaised.

Thapar University cutoff 2025 via JEE Main has been released for B.E./BTech courses in the form of opening and closing ranks. As per the Thapar University JEE Main cutoff 2025, the cutoff rank needed between 46226 and 223330. Similarly, the needed rank for OBC All India quota for one and only course, CSE, is 46226. Similarly, if we talk about the General Home State quota, the needed rank must ranged between 82758 and 501201. To know more cutoff-related information, candidates can click here:

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