A matchbox measures 4 cm × 2.5 cm × 1.5 cm.
Question: A matchbox measures 4 cm 2.5 cm 1.5 cm. What is the volume of a packet containing 12 such matchboxes? Solution: Volume of each matchbox =4 2.5 1.5 = 15 cm3 Volume of 12 matchboxes = 12 15 = 180cm3Thus, the volume of a packet containing 12 such matchboxes is 180cm3....
Read More →be a twice differentiable function such that
Question: Let $f:[0: 2] \rightarrow \mathrm{R}$ be a twice differentiable function such that $f^{\prime \prime}(x)0$, for all $x \in(0,2)$. If $\phi(x)=f(x)+f(2-x)$, then $\phi$ is :(1) increasing on $(0,1)$ and decreasing on $(1,2)$.(2) decreasing on $(0,2)$(3) decreasing on $(0,1)$ and increasing on $(1,2)$.(4) increasing on $(0,2)$Correct Option: , 3 Solution: $\phi(x)=f(x)+f(2-x)$ Now, differentiate w.r.t. $x$, $\phi^{\prime}(x)=f^{\prime}(x)-f^{\prime}(2-x)$ For $\phi(x)$ to be increasing $...
Read More →2,4 -DNP test can be used to identify:
Question: 2,4 -DNP test can be used to identify:aldehydehalogensetheramineCorrect Option: 1 Solution:...
Read More →An emf of 20 V is applied at time t=0
Question: An emf of $20 \mathrm{~V}$ is applied at time $t=0$ to a circuit containing in series $10 \mathrm{mH}$ inductor and $5 \Omega$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40 \mathrm{~s}$ is close to: (Take $e^{2}=7.389$ )(1) $1.06$(2) $1.15$(3) $1.46$(4) $0.84$Correct Option: 1 Solution: (1) The current (I) in LR series circuit is given by $I=\frac{V}{R}\left(1-e^{-\frac{t R}{L}}\right)$ At $t=\infty$, $I_{\infty}=\frac{20}{5}\left(I-e^{\frac{-\infty}{L / R}}\right...
Read More →Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
Question: Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:(i) length = 12 cm, breadth = 8 cm and height = 4.5 cm(ii) length = 26 m, breadth = 14 m and height = 6.5 m(iii) length = 15 m, breadth = 6 m and height = 5 dm(iv) length = 24 m, breadth = 25 cm and height = 6 m Solution: (i)Here,l= 12 cm,b= 8 cm,h= 4.5 cm Volume of the cuboid $=l \times b \times h$ $=(12 \times 8 \times 4.5) \mathrm{cm}^{3}$ $=432 \mathrm{~cm}^{3}$ Total Surface area...
Read More →The major product of the following reaction is :
Question: The major product of the following reaction is : $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$Correct Option: , 3 Solution:...
Read More →respectively the sets of local minimum and local maximum points of the function,
Question: If $S_{1}$ and $S_{2}$ are respectively the sets of local minimum and local maximum points of the function, $f(x)=9 x^{4}+12 x^{3}-36 x^{2}+25, x \in \mathrm{R}$, then :(1) $\mathrm{S}_{1}=\{-2\} ; \mathrm{S}_{2}=\{0,1\}$(2) $\mathrm{S}_{1}=\{-2,0\} ; \mathrm{S}_{2}=\{1\}$(3) $\mathrm{S}_{1}=\{-2,1\} ; \mathrm{S}_{2}=\{0\}$(4) $\mathrm{S}_{1}=\{-1\} ; \mathrm{S}_{2}=\{0,2\}$Correct Option: , 3 Solution: $f(x)=9 x^{4}+12 x^{3}-36 x^{2}+25$ $f^{\prime}(x)=36\left[x^{3}+x^{2}-2 x\right]=3...
Read More →Given below are two statements:
Question: Given below are two statements: Statement-I : $\mathrm{CeO}_{2}$ can be used for oxidation of aldehydes and ketones. Statement-II : Aqueous solution of EuSO $_{4}$ is a strong reducing agent.Statement I is true, statement II is falseStatement $\mathrm{I}$ is false, statement $\mathrm{II}$ is trueBoth Statement I and Statement II are falseBoth Statement I and Statement II are trueCorrect Option: , 4 Solution: $\mathrm{CeO}_{2}$ can be used as oxidising agent like seO $_{2} .$ Similarly ...
Read More →A LCR circuit behaves like a damped harmonic oscillator.
Question: A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ' $b$ ', the correct equivalence would be:(1) $L \leftrightarrow m, C \leftrightarrow k, R \leftrightarrow b$(2) $L \leftrightarrow \frac{1}{b}, C \leftrightarrow \frac{1}{m}, R \leftrightarrow \frac{1}{k}$(3) $L \leftrightarrow k, C \leftrightarrow b, R \leftrightarrow m$(4) $L \leftrightarrow m, C \leftrightarrow \frac{1}{k}, R \leftrightarrow b$...
Read More →Which of the following reagent is suitable for the preparation of the product in the above reaction.
Question: Which of the following reagent is suitable for the preparation of the product in the above reaction.$\operatorname{Red} P+C l_{2}$$\mathrm{Ni} / \mathrm{H}_{2}$$\mathrm{NaBH}_{4}$Correct Option: , 2 Solution: It is wolf-kishner reduction of carbonyl compounds....
Read More →The shortest distance between the
Question: The shortest distance between the line $y=x$ and the curve $y^{2}=x-2$ is :(1) 2(2) $\frac{7}{8}$(3) $\frac{7}{4 \sqrt{2}}$(4) $\frac{11}{4 \sqrt{2}}$Correct Option: , 3 Solution: The shortest distance between line $y=x$ and parabola $=$ the distance $L M$ between line $y=x$ and tangent of parabola having slope 1 . $y=m(x-2)+\frac{a}{m}$ Here $m=1$ and $a=\frac{1}{4}$ $\therefore$ equation of tangent is: $y=x-\frac{7}{4}$ Distance between the line $y-x=0$ and $y-x+\frac{7}{4}=0$ $=\lef...
Read More →Wich one of the following carbonyl compounds cannot be prepared by
Question: Wich one of the following carbonyl compounds cannot be prepared by addition of wate on an alkyne in the presence of $\mathrm{HgSO}_{4}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ ?Correct Option: 1 Solution: Reaction of Alkyne with $\mathrm{HgSO}_{4} \ \mathrm{H}_{2} \mathrm{SO}_{4}$ follow as Hence, by this process preparation of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$ Cann't possible....
Read More →Wich one of the following carbonyl compounds cannot be prepared by
Question: Wich one of the following carbonyl compounds cannot be prepared by addition of wate on an alkyne in the presence of $\mathrm{HgSO}_{4}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ ?Correct Option: 1 Solution: Reaction of Alkyne with $\mathrm{HgSO}_{4} \ \mathrm{H}_{2} \mathrm{SO}_{4}$ follow as Hence, by this process preparation of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$Cann't possible....
Read More →In a series LR circuit, power of 400W is dissipated from a source of 250 V
Question: In a series LR circuit, power of $400 \mathrm{~W}$ is dissipated from a source of $250 \mathrm{~V}, 50 \mathrm{~Hz}$. The power factor of the circuit is $0.8$. In order to bring the power factor to unity, a capacitor of value $\mathrm{C}$ is added in series to the $\mathrm{L}$ and $\mathrm{R}$. Taking the value $\mathrm{C}$ as $\left(\frac{\mathrm{n}}{3 \pi}\right) \mu \mathrm{F}$, then value of $\mathrm{n}$ is Solution: $(400)$ Given: Power $P=400 \mathrm{~W}$, Voltage $V=250 \mathrm{...
Read More →Which of the following reagent is used for the following reaction?
Question: Which of the following reagent is used for the following reaction? $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} \stackrel{?}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$Manganese acetateCopper at high temperature and pressureMolybdenum oxidePotassium permanganateCorrect Option: , 2 Solution:...
Read More →The area of an equilateral triangle is
Question: The area of an equilateral triangle is $81 \sqrt{3} \mathrm{~cm}^{2}$. Its height is (a) $9 \sqrt{3} \mathrm{~cm}$ (b) $6 \sqrt{3} \mathrm{~cm}$ (c) $18 \sqrt{3} \mathrm{~cm}$ (d) 9cm Solution: (a) $9 \sqrt{3} \mathrm{~cm}$ Area of equilateral triangle $=81 \sqrt{3} \mathrm{~cm}^{2}$ $\Rightarrow \frac{\sqrt{3}}{4} \times(\text { Side })^{2}=81 \sqrt{3}$ $\Rightarrow(\text { Side })^{2}=81 \times 4$ $\Rightarrow(\text { Side })^{2}=324$ $\Rightarrow$ Side $=18 \mathrm{~cm}$ Now, Height...
Read More →Let a function f:[0,5]
Question: Let a function $f:[0,5] \rightarrow \mathbf{R}$ be continuous, $f(1)=3$ and $\mathrm{F}$ be defined as: $\mathrm{F}(x)=\int_{1}^{x} t^{2} g(t) d t$, where $g(t)=\int_{1}^{x} f(u) d u$ Then for the function $\mathrm{F}$, the point $x=1$ is : (1) a point of local minima.(2) not a critical point.(3) a point of local maxima.(4) a point of inflection.Correct Option: 1 Solution: $F(x)=\int_{1}^{x} t^{2} g(t) d t$ Differentiate by using Leibnitz's rule, we get $F^{\prime}(x)=x^{2} g(x)=x^{2} ...
Read More →In Tollen's test for aldehyde,
Question: In Tollen's test for aldehyde, the overall number of electron(s) transferred to the Tollen's reagent formula $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$per aldehyde group to form silver mirror is . _____________(Round off to the Nearest integer) Solution: (2) Total $2 \mathrm{e}^{-}$transfer to Tollen's reagent...
Read More →The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long.
Question: The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is(a) 168 cm2(b) 252 cm2(c) 336 cm2(d) 504cm2 Solution: (c) 336 cm2 Let $\triangle P Q R$ be a right-angled triangle and $\mathrm{PQ} \perp \mathrm{QR}$. Now, $P Q=\sqrt{P R^{2}-Q R^{2}}$ $=\sqrt{50^{2}-48^{2}}$ $=\sqrt{2500-2304}$ $=\sqrt{196}$ $=14 \mathrm{~cm}$ $\therefore$ Area of triangle $=\frac{1}{2} \times Q R \times P Q=\frac{1}{2} \times 48 \times 14=336 \mathrm{~cm}^{2}$...
Read More →In Tollen's test for aldehyde,
Question: In Tollen's test for aldehyde, the overall number of electron(s) transferred to the Tollen's reagent formula $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$per aldehyde group to form silver mirror is . _____________(Round off to the Nearest integer) Solution: (2) $\mathrm{AgNO}_{3}+\mathrm{NaOH} \rightarrow \mathrm{AgOH}+\mathrm{NaNO}_{3}$ $2 \mathrm{AgOH} \rightarrow \mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}$ $\mathrm{Ag}_{2} \mathrm{O}+4 \mathrm{NH}_{3}+\mat...
Read More →Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
Question: Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is(a) 156 cm2(b) 78 cm2(c) 60 cm2(d) 120cm2 Solution: (c) $60 \mathrm{~cm}^{2}$ Area of isosceles triangle $=\frac{b}{4} \sqrt{4 a^{2}-b^{2}}$ Here, $a=13 \mathrm{~cm}$ and $b=24 \mathrm{~cm}$ Thus, we have : $\frac{24}{4} \times \sqrt{4(13)^{2}-24^{2}}$ $=6 \times \sqrt{676-576}$ $=6 \times \sqrt{100}$ $=6 \times 10$ $=60 \mathrm{~cm}^{2}$...
Read More →A spherical iron ball of
Question: A spherical iron ball of $10 \mathrm{~cm}$ radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \mathrm{~cm}^{3} /$ min. When the thickness of ice is $5 \mathrm{~cm}$, then the rate (in $\mathrm{cm} / \mathrm{min} .)$ at which of the thickness of ice decreases, is:(1) $\frac{5}{6 \pi}$(2) $\frac{1}{54 \pi}$(3) $\frac{1}{36 \pi}$(4) $\frac{1}{18 \pi}$Correct Option: , 4 Solution: Let the thickness of ice layer be $=x \mathrm{~cm}$ Total volume $V=\frac{4...
Read More →A part of a complete circuit is shown in the figure. At some
Question: A part of a complete circuit is shown in the figure. At some instant, the value of current $\mathrm{I}$ is $1 \mathrm{~A}$ and it is decreasing at a rate of $10^{2} \mathrm{~A} \mathrm{~s}^{-1}$. The value of the potential difference $\mathrm{V}_{\mathrm{P}}$ $-\mathrm{V}_{\mathrm{Q}}$, (in volts) at that instant, is Solution: (33) Here, $L=50 \mathrm{mH}=50 \times 10^{-3} \mathrm{H} ; I=1 \mathrm{~A}, R=2 \Omega$ $V_{P}-L \frac{d l}{d t}-30+R I=V_{Q}$ $\Rightarrow V_{P}-V_{Q}=50 \time...
Read More →The area of an equilateral triangle is
Question: The area of an equilateral triangle is $36 \sqrt{3} \mathrm{~cm}^{2}$. Its perimeter is (a) 36 cm (b) $12 \sqrt{3} \mathrm{~cm}$ (c) 24 cm(d) 30cm Solution: (a) 36 cm Area of equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$ $\Rightarrow \frac{\sqrt{3}}{4} \times(\text { Side })^{2}=36 \sqrt{3}$ $\Rightarrow(\text { Side })^{2}=144$ $\Rightarrow$ Side $=12 \mathrm{~cm}$ Now,Perimeter=3Side=312=36cm...
Read More →Consider the above reaction,
Question: Consider the above reaction, the product ' $\mathrm{X}$ ' and 'Y' respectively are : Correct Option: , 3 Solution:...
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