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The ranges and heights for two projectiles projected ....

The ranges and heights for two projectiles projected with the same initial velocity at angles $42^\circ $ and $48^\circ $ with the horizontal are $R_1\,,R_2$ and $H_1\,,H_2$ respectively. Choose the correct option: (A) $R_1 < R_2 $ and $H_1 < H_2 $ (B) $R_1 > R_2 $ and $H_1 = H_2 $ (C) $R_1 = R_2 $ and $H_1 = H_2 $ (D) $R_1 = R_2 $ and $H_1 < H_2 $ Solution $R=\frac {u^2 sin 2\theta }{g} $, $H=\frac {u^2 sin^2 \theta }{2g}$  $\frac {R_1}{R_2} = \frac {sin 2 \times 42^\circ}{sin 2\times 48^\circ} = \frac {cos 6^\circ}{cos 6^\circ} = 1 $ $\frac {H_1}{H_2}= \frac {sin^2 42^\circ }{sin^2 48^\circ } = tan^2 42^\circ < 1 $ Answer: (D)
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The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)=?$

The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is: (A) $2\sqrt 6$ (B) $\sqrt {26}$ (C) $2\sqrt 5 $ (D) $4\sqrt 2 $ Solution We have, $3x-z+4=0$ or $z=3x+4$ & $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$ Any point $(x,y,z)$ on the line $ \equiv (t,2t + 3,3t + 4)$ Let $d$ be the distance between $(2,-1,6)$ & $ (t,2t + 3,3t + 4)$ Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$ Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0. Answer: (A)

$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$

$cos^{-1}cos(-5)+$

$sin^{-1}sin6$-

$tan^{-1}tan12 =? $

$cos^{-1}(cos(-5))+sin^{-1}(sin(6))-tan^{-1}(tan(12)) $ is equal to:  (The inverse trigonometric functions take the principal values) (A) $3\pi - 11$ (B) $3\pi + 1 $ (C) $4\pi - 11$ (D) $4\pi - 9$ Solution The given expression can be rewritten as, $cos^{-1}cos5+sin^{-1}sin 6-tan^{-1} tan12 $ Considering the principal values the given expression can be further rewritten as, $cos^{-1}cos(2\pi - 5)+sin^{-1}sin (-(2\pi- 6))-tan^{-1} tan(-(4\pi -12)) $ Or $(2\pi - 5)+ (-(2\pi- 6))-(-(4\pi-12)) = 4\pi -11 $ Answer: (C)

A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$

Planes $x-2y-2z+1=0$ & $2x-3y-6z+1=0$ ....

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P? (A) $(0,2,-4)$ (B) $(4,0,-2)$ (C) $(-2,0,-\frac {1}{2})$ (D) $(3,1,-\frac {1}{2})$ Solution Bisectors are given by, $\frac{{x - 2y - 2z + 1}}{3} =  \pm \frac{{2x - 3y - 6z + 1}}{7}$ $ \Rightarrow 7x - 14y - 14z + 7 =  \pm (6x - 9y - 18z + 3)$ Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$ Let $\theta $ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$ $\cos \theta  = \frac{{|1 + 10 - 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$ $\theta > 45^\circ $ So, $x-5y+4z+4=0$ is the obtuse angle bisector. $\therefore $ Acute angle bisector $P \equiv 13x - 23y - 32z + 10 = 0$ The point $(-2,0,-\frac {1}{2})$ satisfies P. Answer: (C)

80 g of copper sulphate $CuSO_4 .5H_2 O $ ....

80 g of copper sulphate $CuSO_4 .5H_2 O $ is dissolved in deionised water to make 5 L of solution. The concentration of the copper sulphate solution is $x \times 10^{-3} mol L^{-1} $. The value of x is _ _ _ _ . (Nearest integer) [Atomic masses - Cu:63.54 u, S:32 u, O: 16 u, H: 1 u] Solution $Molarity = \frac {n}{V} = \frac {w}{M_0 L} $ $\therefore Molarity = \frac {80}{249.54 \times 5 } \approx 64 \times 10^{-3} mol L^{-1} $ $\therefore x = 64 $

For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g) $ ....

For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g) $, when $\Delta S = -176.0 JK^{-1} $ and $\Delta H = -57.8 kJ mol^{-1} $, the magnitude of $\Delta G $ at 298 K for the reaction is _ _ _ _ $kJ mol^{-1} $. (Nearest integer) Solution We have, $\Delta G = \Delta H - T \Delta S $ $\Delta S = -176.0 JK^{-1} = -0.176 kJ K^{-1} $ $\Delta G =  -57.8 - 298 \times (-0.176) \approx -57.8+52.5 = -5.3 $ $\therefore |\Delta G | = 5.3  kJ mol^{-1} $ Ans: 5

The molar solubility of $Zn(OH)_2 $ ....

The molar solubility of $Zn(OH)_2 $ in 0.1 M NaOH solution is $x \times 10^{-18} M$. The value of x is _ _ _ _ . (Nearest integer) (Given: The solubility product of $Zn(OH)_2 $ is $2 \times 10^{-20} $) Solution $Zn{(OH)_2}\rightleftharpoons \mathop {Z{n^{2 + }}}\limits_s  + 2\mathop {O{H^ - }}\limits_{2s + 0.1} $ ${K_{sp}} = [Z{n^{2 + }}]{[O{H^ - }]^2} = s.{(2s + 0.1)^2}$ $\therefore 2 \times {10^{ - 20}} = s.{(2s + 0.1)^2} \approx {0.1^2} \times s$ $ \Rightarrow s = 2 \times {10^{ - 18}}$ $\therefore x = 2 $

An empty LPG cylinder weighs 14.8 kg ....

An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside the cylinder is _ _ _ _ atm. (Nearest integer) (Assume LPG to be an ideal gas) Solution Weight of gas before use = 29.0 - 14.8 = 14.2 kg Weight of gas after use = 23.0 - 14.8 = 8.2 kg Now, $\frac {P_1}{n_1} = \frac {P_2}{n_2} $ $\therefore \frac {3.47}{14.2} =  \frac {P_2}{8.2} $ $\Rightarrow P_2 = \frac {3.47 \times 8.2}{14.2} \approx 2 atm $ Answer: 2

If the conductivity of mercury at $0^\circ C $ is ....

If the conductivity of mercury at $0^\circ C $ is $1.07 \times 10^6 Sm^{-1} $ and the resistance of a cell containing mercury is $0.243 \Omega $, then the cell constant of the cell is $x \times 10^4 m^{-1} $. (Nearest integer) Solution $R=\frac {K_{cell}}{\kappa }$ Or, $K_{cell} = 0.243 \times 1.07 \times 10^6 \approx 26 \times 10^4 m^{-1} $ Answer: 26

$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy} $

f(x)=?

The function f(x), that satisfies the condition $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy} $ is: (A) $x + \frac{2}{3}(\pi  - 2)\sin x$ (B) $x + (\pi  + 2)\sin x$ (C) $x + \frac{\pi }{2}\sin x$ (D) $x + (\pi  - 2)\sin x$ Solution We have, $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy}  = x + \sin x\int\limits_0^{\pi /2} {\cos yf(y)dy}  = x + \sin x.k$ Where, $k = \int\limits_0^{\pi /2} {\cos yf(y)dy}  = \int\limits_0^{\pi /2} {\cos y(y + \sin y.k)dy} $ $\therefore k = \int\limits_0^{\pi /2} {y\cos ydy}  + k\int\limits_0^{\pi /2} {\cos y\sin ydy} $ $ \Rightarrow k = \left. {y\sin y} \right|_0^{\pi /2} - \int\limits_0^{\pi /2} {\sin ydy}  + \frac{k}{2}\int\limits_0^{\pi /2} {sin2ydy}  = \frac{\pi }{2} + \left. {\cos y} \right|_0^{\pi /2} - \frac{k}{2}.\frac{1}{2}\left. {\cos 2y} \right|_0^{\pi /2}$ $ \Rightarrow k = \frac{\pi }{2} - 1 - \frac{k}{4}( - 1 - 1) = \frac{\pi }{2} - 1 + \frac{k}{2}$ $ \Rightarrow \frac{k}{2} = \frac{\pi }{2} - 1 = \frac{{\pi  - 2}}{2}$ $

A 2 kg steel rod of length 0.6 m is clamped ....

A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is _ _ _ _ $m s^{-1}$. Solution Decrease in gravitational potential energy = Increase in kinetic energy So, $mgl = \frac {1}{2} I \omega ^2 = \frac {1}{2} .\frac {1}{3} ml^2 \omega ^2 $ $\Rightarrow 6g = l \omega ^2 $ Using $v = l \omega $, we have $6 g = l. \frac {v^2}{l^2}$ $\Rightarrow v^2 = 6gl $ $\Rightarrow v = \sqrt {6gl} = \sqrt {6\times 10 \times 0.6 } = 6ms^{-1} $ Answer: 6

A cube is placed inside an electric field, $\vec E = 150 y^2 \hat j $ ....

A cube is placed inside an electric field, $\vec E = 150 y^2 \hat j $. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is: (A) $8.3 \times 10^{-11} C$ (B) $8.3 \times 10^{-12} C$ (C) $3.8 \times 10^{-12} C$ (D) $3.8 \times 10^{-11} C$ Solution We have, $\phi = \frac {q_{in}}{\epsilon_0}$ Since electric field is upwardly directed, it will only cut the top and bottom surfaces. Since y = 0 at the bottom surface makes electric field 0 there, no flux is present there. The only flux through the cube is contributed by the flux through the top surface. Flux through the top surface $=150 \times 0.5^2 \times 0.5^2 $ So, $\phi =150 \times 0.5^2 \times 0.5^2 = \frac {q_{in}}{8.85 \times 10^{-12}}$ $\therefore q_{in} \approx 8.3 \times 10^{-11} C $ Answer: (A)

A glass tumbler having inner depth of 17.5 cm ....

A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water $(\mu = 4/3 )$ into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled? (A) 11.7 cm (B) 8.75 cm (C) 7.5 cm (D) 10 cm Solution $\mu = \frac {Real\, Depth}{Apparent\, Depth} = \frac {17.5-y}{y}$ $\therefore \frac {4}{3} = \frac {17.5-y}{y} $ $\therefore y = 7.5 cm$ Real Depth = 17.5 - y = 10 cm Answer: (D)

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)

An engine is attached to a wagon through a shock absorber ....

An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of $72 kmh^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, the spring constant is _ _ _ _ $\times 10^5 Nm^{-1}$. Solution 10 % of K.E. = $\frac {1}{2} kx^2 $ $\therefore 0.1 \times \frac {1}{2} mv^2 = \frac {1}{2} kx^2 $ $\therefore 0.1 \times 40,000 \times (72 \times \frac {5}{18})^2 = k \times 1.0^2 $ $\Rightarrow k = 4,000 \times 20^2 = 16 \times 10^5 Nm^{-1} $ Answer: 16

The average translational kinetic energy of $N_2 $ ....

The average translational kinetic energy of $N_2 $ gas molecules at _ _ _ _ $^\circ C $ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 Volt. (Given $k_b = 1.38 \times 10^{-23} J/K $) (Fill the nearest integer) Solution K.E. of electron =  Translational K.E. of $N_2 $ molecules $\therefore eV = \frac {3}{2} kT $ $\therefore 2 \times 1.6 \times 10^{-19} \times 0.1 = 3 \times 1.38 \times 10^{-23} \times T $ $\therefore T = \frac {1.6 \times 10^3 }{3 \times 0.69 } \approx 773 K = 500 ^\circ C $ Answer: 500