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\(5^{2(x - 1)} \times 5^{x + 1} = 0.04\) \(5^{2x - 2} \times 5^{x + 1} = 5^{-2}\) \(2x - 2 + x + 1 = -2\) \(3x - 1 = -2 \implies 3x = -2 + 1 = -1\) \(x = -\frac{1}{3}\)

log (\(\frac{28}{3}) = \log (\frac{7 \times 4}{3})\) = log 7 + log 4 - log 3 log 7 + log \(2^2\) - log 3 = log 7 + 2log2 - log 3 = z + 2x - y = 2x + z - y

x = \(\frac{-1}{2}\) x = 3 2x = -1, x = 3; 2x + 1 = 0 x - 3 = 0 (2x + 1)(x - 3) = 0 2x2 - 6x + x - 3 = 0 2x2 - 5x - 3 = 0

(2x + 5)o + (xo + 20o) + x + (3x - 20)o + (x + 15)o = (n - 2) x 180 8n + 20 = (5 - 2) x 180 where n = 5(Pentagon) 8n + 20 = 3 x 180 8n + 20 = 540 8n = 540 - 20 8n = 520 x = 65o

x(4 - x) 4x - x2 \(\frac{dy}{dx}\) = 4 - 2x \(\frac{dy}{dx}\) = 0 2x = 4 x = \(\frac{4}{2}\) = 2

\(\begin{array}{c|c} Mark & No. of Candidate\\\hline 60 & 40\\ 70 & 25\\ 80 & 15\\ 90 & 10 \\100 & 5 \\\hline & 95 \end{array}\)

the 36th percentile i.e. 36% of the total frequency of 50. \(\frac{36}{100}\) \(\times\) 50 = 18 = From the graph, 18 on the vertical axis corresponds to28 on the horizontal axis. \(\therefore\) P\(_{36}\) = 28%

A pyrometer is instrument for measuring high temperature. The optical pyrometer is one in which the luminous radiation from the hot body is compared with that from a known source.

Heat supplied = Heat gained pt = ml T = \(\frac{ml}{p}\) = \(\frac{2 \times 2.3 \times 10^6}{1000}\) = 4.6 x 10\(^3\) sec

Radiation is the method of heat transfer with or without a material medium. It can be through space or through a material medium.

Acceleration = \(\frac{\DeltaV}{\Deltat}\) = \(\frac{V - U}{t_2 - t_1}\) = \(\frac{35 - 15}{20 - 10}\) = \(\frac{20}{10}\) = 2ms\(^-2\)

From S = ut + 1/2 at2 S = 1/2 ut2, since U = 0 Let S5 = distance in 5 seconds = 1/2 x 4 x 52 = 2 x 25 = 50m S4 = distance covered in 4 seconds = 1/2 x 4 x 42 = 2 x 16 = 32m Distance covered in the fifth second S5 - S4 = 50m - 32m = 18m

Power = \(\frac{work done}{time taken}\) where 1 min = 60s = \(\frac{mgh}{t}\) = \(\frac{750 \times 10 \times 90}{60}\) = 11250w or 11.25kw

coefficient of friction = \(\frac{F}{R}\) =\(\frac{F}{mg}\) = \(\frac{500}{80 \times 10}\) = 0.625

V.R = \(\frac{1}{sin\theta}\) = \(\frac{1}{sin 30^\circ}\) = \(\frac{1}{0.5}\) = 2