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\(^{10}C_6 = \frac{10!}{(10-6)!6!}\) = \(\frac{10!}{4!6!}\) =\(\frac{10\times 9\times 8 \times 7 \times 6!}{4\times 3\times 2\times 1 \times 6!}\) = 210.

OBSTRUCTION Total possible outcome = 11 Number of chance of getting T = 2 P(picking T) = 2/11

Total possible outcome 12+18+x+30+2x+45 = 105+3x ∴105+3x = 150 3x = 150-105 3x = 45 x = 15 P(obtaining 5) = \(\frac{2x}{(105+3x)}\) But x= 15 =\(\frac{2(15)}{(105+3(15))}\) =\(\frac{30}{(105+45)}\) =\(\frac{30}{150}\) =\(\frac{1}{5}\)

\(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\) therefore, (x, y) = (-3, -5) respectively

-2p + r = 1.......(i) 2p + 3r = 0.......(ii) r - 1 + 2p ........(iii) 2p + 3(1 + 2p) = 0 2p + 3(1 + 2p) = 0 2p + 3 + 6p = 0 3 - 8p = 0 \(\to\) 8p = 3 p = \(\frac{3}{8}\) 6 = 1 - 2 \(\frac{3}{8}\) = 1 - \(\frac{6}{8}\) \(\frac{8 - 6}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)

L\(\begin{pmatrix} x_1 & y_1 \\ -1 & -6 \end{pmatrix}\) m L\(\begin{pmatrix} x_2 & y_2 \\ -3 & -5 \end{pmatrix}\) D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) D = \(\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2}\) D = \(\sqrt{(-3 + 1)^2 + (-5 + 6)^2}\) D = \(\sqrt{(-2)^2 + 1^2}\) D = \(\sqrt{4 + 1}\) D = \(\sqrt{5}\)

The silver coating on the inner bottle of a thermos flask is used to prevent heat transfer by radiation

V = wr = 2\pif r = 2 x (22)/7 x 1 x 0.6 = 3.77ms-1

The friction between the tanker and air creates charges which the chain conducts to the ground.

The force which is centripetal F = \(\frac{MV^2}{R}\) = \(\frac{5 \times 7^2}{3.5}\) = 70 Since force is 70 N when speed is 7m/s Speed = ? when force 70N is tripled → 70 * 3 = 210N 210 = \(\frac{5\timesv^2}{3.5}\) 210 \times 3.5 = 5 * v\(^2\) 735 \div 5 = v\(^2\) v\(^2\) = 147 v =12.12m/s

Smooth plane has negligible friction ∴ only the resultant of the perpendicular forces is acting on the body, which is given as F = \(\sqrt{ 6^2 + 8^2}\) = \(\sqrt{64 + 36}\) = \(\sqrt{100}\) = 10N But F = Ma ∴ a = F/M = 10/4 = 2.5 ms-1

VR = No. of pulleys = 4 Efficiency = \(\frac{M.A}{V.R}\) cross multiply \(\frac{80}{100} * V.R\) = M.A \(\frac{80}{100} * 4\) = M.A M.A = 3.25 where M.A = \(\frac{load}{effort}\) ∴ Effort = \(\frac{1200}{3.2} * V.R\) = 375N

The condition (ii) is a condition for an equilibrium of parallel forces only while conditions (i) and (iii) are correct for both parallel and non- parallel coplanar forces. No specifications are made, so all conditions apply.

If 2P is added P, the beam will bend at R side. So to keep it in equilibrium. P should be moved nearer to 0

Frequency = (No. of revolutions)/ Time = (90)/25 = 3.6 Hz