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Question Number 23431 Answers: 0 Comments: 0

If X is a discrete random variable then p(X≥a)=

$${If}\:{X}\:{is}\:{a}\:{discrete}\:{random}\:{variable}\:{then}\:{p}\left({X}\geqslant{a}\right)= \\ $$

Question Number 23429 Answers: 0 Comments: 1

An asymptote to the curve y^2 (1+x)=x^2 (1−x) is

$${An}\:{asymptote}\:{to}\:{the}\:{curve}\:{y}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)={x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\:{is} \\ $$

Question Number 23428 Answers: 2 Comments: 0

the curve ay^2 =x^2 (3a−x) cuts the y-axis at

$${the}\:{curve}\:{ay}^{\mathrm{2}} ={x}^{\mathrm{2}} \left(\mathrm{3}{a}−{x}\right)\:{cuts}\:{the}\:{y}-{axis}\:{at} \\ $$

Question Number 23426 Answers: 1 Comments: 0

just a silly question: write a correct maths equation using only symbols below. Each must be used and only once. 2, 3, 4, 5, =, +

$${just}\:{a}\:{silly}\:{question}: \\ $$$${write}\:{a}\:{correct}\:{maths}\:{equation} \\ $$$${using}\:{only}\:{symbols}\:{below}.\:{Each} \\ $$$${must}\:{be}\:{used}\:{and}\:{only}\:{once}. \\ $$$$\:\:\:\:\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:=,\:+ \\ $$

Question Number 23425 Answers: 0 Comments: 0

a,b,c>0 ⇒((a^3 +b^3 +c^3 )/((a+b)(b+c)(c+a)))≥(3/8) ?

$$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0}\:\Rightarrow\frac{\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}} +\boldsymbol{{c}}^{\mathrm{3}} }{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{c}}+\boldsymbol{{a}}\right)}\geqslant\frac{\mathrm{3}}{\mathrm{8}}\:? \\ $$

Question Number 23419 Answers: 2 Comments: 0

Question Number 23418 Answers: 1 Comments: 0

∫sec^2 (√x) /(√x) dx

$$\int\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{x}}\:/\sqrt{\mathrm{x}}\:\mathrm{dx} \\ $$

Question Number 23411 Answers: 2 Comments: 1

Question Number 23409 Answers: 1 Comments: 0

Hybridisation of N in HNO_3 is

$$\mathrm{Hybridisation}\:\mathrm{of}\:\mathrm{N}\:\mathrm{in}\:\mathrm{HNO}_{\mathrm{3}} \:\mathrm{is} \\ $$

Question Number 23408 Answers: 1 Comments: 0

Question Number 23402 Answers: 0 Comments: 2

Question Number 23394 Answers: 1 Comments: 0

Which is correct statement? (1) The entropy of the universe increases and tends towards the maximum value (2) All natural processes are irreversible (3) For reversible isolated processes, change of entropy is zero (4) For irreversible expansion of isolated processes, entropy change < 0

$$\mathrm{Which}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{statement}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{entropy}\:\mathrm{of}\:\mathrm{the}\:\mathrm{universe}\:\mathrm{increases} \\ $$$$\mathrm{and}\:\mathrm{tends}\:\mathrm{towards}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{value} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{natural}\:\mathrm{processes}\:\mathrm{are}\:\mathrm{irreversible} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{For}\:\mathrm{reversible}\:\mathrm{isolated}\:\mathrm{processes}, \\ $$$$\mathrm{change}\:\mathrm{of}\:\mathrm{entropy}\:\mathrm{is}\:\mathrm{zero} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{For}\:\mathrm{irreversible}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\mathrm{isolated}\:\mathrm{processes},\:\mathrm{entropy}\:\mathrm{change}\:<\:\mathrm{0} \\ $$

Question Number 23393 Answers: 0 Comments: 1

Show that volume of a region of space bounded by a boundary surface S is V= (1/3)∫∫_(S ) rcos θdA . θ being the angle between the position vector of a point P on the surface, and the outer normal to the surface at P. r is the distance of point P from origin.

$${Show}\:{that}\:{volume}\:{of}\:{a}\:{region} \\ $$$${of}\:{space}\:{bounded}\:{by}\:{a}\:{boundary} \\ $$$${surface}\:{S}\:{is}\:\:{V}=\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{S}\:} {\int\int}{r}\mathrm{cos}\:\theta{dA}\:. \\ $$$$\theta\:{being}\:{the}\:{angle}\:{between}\:{the} \\ $$$${position}\:{vector}\:{of}\:{a}\:{point}\:{P}\:\:{on} \\ $$$${the}\:{surface},\:{and}\:{the}\:{outer}\:{normal} \\ $$$${to}\:{the}\:{surface}\:{at}\:{P}. \\ $$$${r}\:{is}\:{the}\:{distance}\:{of}\:{point}\:{P}\:{from} \\ $$$${origin}. \\ $$

Question Number 23390 Answers: 1 Comments: 2

Question Number 23399 Answers: 1 Comments: 4

Two blocks of masses 2 kg and 3 kg are kept on a smooth inclined plane. A constant force of magnitude 20 N is applied on 2 kg block parallel to the inclined. The contact force between the two blocks is

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{3}\:\mathrm{kg} \\ $$$$\mathrm{are}\:\mathrm{kept}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{inclined}\:\mathrm{plane}. \\ $$$$\mathrm{A}\:\mathrm{constant}\:\mathrm{force}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{20}\:\mathrm{N}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{inclined}.\:\mathrm{The}\:\mathrm{contact}\:\mathrm{force}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{blocks}\:\mathrm{is} \\ $$

Question Number 23381 Answers: 1 Comments: 0

A women swimming upstream is not moving with respect to the ground. Is she doing any work, if she stops swimming and merely floats is work done on her?

$$\mathrm{A}\:\mathrm{women}\:\mathrm{swimming}\:\mathrm{upstream}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{Is} \\ $$$$\mathrm{she}\:\mathrm{doing}\:\mathrm{any}\:\mathrm{work},\:\mathrm{if}\:\mathrm{she}\:\mathrm{stops} \\ $$$$\mathrm{swimming}\:\mathrm{and}\:\mathrm{merely}\:\mathrm{floats}\:\mathrm{is}\:\mathrm{work} \\ $$$$\mathrm{done}\:\mathrm{on}\:\mathrm{her}? \\ $$

Question Number 23385 Answers: 0 Comments: 1

Question Number 23369 Answers: 1 Comments: 0

Which of the following pair would correct inequality for standard molar entropy? (1) NO(g) < NO_2 (g) (2) C_2 H_2 (g) > C_2 H_6 (g) (3) CH_3 COOH (l) < HCOOH (l) (4) CO_2 (g) < CO(g)

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{pair}\:\mathrm{would} \\ $$$$\mathrm{correct}\:\mathrm{inequality}\:\mathrm{for}\:\mathrm{standard}\:\mathrm{molar} \\ $$$$\mathrm{entropy}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{NO}\left(\mathrm{g}\right)\:<\:\mathrm{NO}_{\mathrm{2}} \left(\mathrm{g}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{C}_{\mathrm{2}} \mathrm{H}_{\mathrm{2}} \left(\mathrm{g}\right)\:>\:\mathrm{C}_{\mathrm{2}} \mathrm{H}_{\mathrm{6}} \left(\mathrm{g}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{CH}_{\mathrm{3}} \mathrm{COOH}\:\left(\mathrm{l}\right)\:<\:\mathrm{HCOOH}\:\left(\mathrm{l}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{CO}_{\mathrm{2}} \left(\mathrm{g}\right)\:<\:\mathrm{CO}\left(\mathrm{g}\right) \\ $$

Question Number 23368 Answers: 0 Comments: 0

Assertion: The first ionization energy of Be is greater than that of B. Reason: 2p-orbital is lower in energy than 2s-orbital.

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{The}\:\mathrm{first}\:\mathrm{ionization}\:\mathrm{energy} \\ $$$$\mathrm{of}\:\mathrm{Be}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{that}\:\mathrm{of}\:\mathrm{B}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{2}{p}-\mathrm{orbital}\:\mathrm{is}\:\mathrm{lower}\:\mathrm{in}\:\mathrm{energy} \\ $$$$\mathrm{than}\:\mathrm{2}{s}-\mathrm{orbital}. \\ $$

Question Number 23364 Answers: 0 Comments: 0

Mode=17, mean=10.22, median=10 Describe the shape of distribution.

$${Mode}=\mathrm{17},\:{mean}=\mathrm{10}.\mathrm{22},\:{median}=\mathrm{10} \\ $$$${Describe}\:{the}\:{shape}\:{of}\:{distribution}. \\ $$

Question Number 23362 Answers: 0 Comments: 0

Question Number 23361 Answers: 1 Comments: 0

A 50 g lead bullet, sp. heat 0.02 is initially at 30°C. It is fired vertically upwards with a speed of 840 m/s and on returning to the starting level strikes a cake of ice at 0°C. How much ice is melted? Assume that all energy is spent in melting only (L = 80 cal/g).

$$\mathrm{A}\:\mathrm{50}\:\mathrm{g}\:\mathrm{lead}\:\mathrm{bullet},\:\mathrm{sp}.\:\mathrm{heat}\:\mathrm{0}.\mathrm{02}\:\mathrm{is} \\ $$$$\mathrm{initially}\:\mathrm{at}\:\mathrm{30}°\mathrm{C}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{vertically} \\ $$$$\mathrm{upwards}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{840}\:\mathrm{m}/\mathrm{s}\:\mathrm{and} \\ $$$$\mathrm{on}\:\mathrm{returning}\:\mathrm{to}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{level}\:\mathrm{strikes} \\ $$$$\mathrm{a}\:\mathrm{cake}\:\mathrm{of}\:\mathrm{ice}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}.\:\mathrm{How}\:\mathrm{much}\:\mathrm{ice}\:\mathrm{is} \\ $$$$\mathrm{melted}?\:\mathrm{Assume}\:\mathrm{that}\:\mathrm{all}\:\mathrm{energy}\:\mathrm{is} \\ $$$$\mathrm{spent}\:\mathrm{in}\:\mathrm{melting}\:\mathrm{only}\:\left({L}\:=\:\mathrm{80}\:\mathrm{cal}/\mathrm{g}\right). \\ $$

Question Number 23358 Answers: 0 Comments: 0

Question Number 23357 Answers: 0 Comments: 0

Prove that (1/(m!))C_0 +(n/((m+1)!))C_1 +((n(n−1))/((m+2)!))C_2 +...+((n(n−1)...2.1)/((m+n)!))C_n = (((m+n+1)(m+n+2)...(m+2n))/((m+n)!)).

$${Prove}\:{that}\:\frac{\mathrm{1}}{{m}!}{C}_{\mathrm{0}} +\frac{{n}}{\left({m}+\mathrm{1}\right)!}{C}_{\mathrm{1}} +\frac{{n}\left({n}−\mathrm{1}\right)}{\left({m}+\mathrm{2}\right)!}{C}_{\mathrm{2}} \\ $$$$+...+\frac{{n}\left({n}−\mathrm{1}\right)...\mathrm{2}.\mathrm{1}}{\left({m}+{n}\right)!}{C}_{{n}} = \\ $$$$\frac{\left({m}+{n}+\mathrm{1}\right)\left({m}+{n}+\mathrm{2}\right)...\left({m}+\mathrm{2}{n}\right)}{\left({m}+{n}\right)!}. \\ $$

Question Number 23348 Answers: 0 Comments: 1

Question Number 23346 Answers: 1 Comments: 7

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