Question and Answers Forum

AllQuestion and Answers: Page 1810

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

Question Number 23334 Answers: 1 Comments: 3

Question Number 23343 Answers: 0 Comments: 2

Question Number 23330 Answers: 0 Comments: 0

The number of ordered pair(s) (x, y) satisfying the equations sinx.cosy = 1 and x^2 + y^2 ≤ 9π^2 is/are

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{pair}\left(\mathrm{s}\right)\:\left({x},\:{y}\right) \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{sin}{x}.\mathrm{cos}{y}\:=\:\mathrm{1} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:\leqslant\:\mathrm{9}\pi^{\mathrm{2}} \:\mathrm{is}/\mathrm{are} \\ $$

Question Number 23317 Answers: 1 Comments: 0

∫sin^3 x cos x dx

$$\int\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{dx} \\ $$

Question Number 23312 Answers: 1 Comments: 2

Question Number 23323 Answers: 0 Comments: 1

Question Number 23309 Answers: 0 Comments: 0

Question Number 23304 Answers: 1 Comments: 2

If sin^4 x + cos^4 y + 2 = 4 sinx.cosy & 0 ≤ x,y ≤ (π/2) , then the value of sinx + cosy is equal to

$$\mathrm{If}\:\mathrm{sin}^{\mathrm{4}} \mathrm{x}\:+\:\mathrm{cos}^{\mathrm{4}} \mathrm{y}\:+\:\mathrm{2}\:=\:\mathrm{4}\:\mathrm{sinx}.\mathrm{cosy}\:\& \\ $$$$\mathrm{0}\:\leqslant\:{x},{y}\:\leqslant\:\frac{\pi}{\mathrm{2}}\:,\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}{x}\:+ \\ $$$$\mathrm{cos}{y}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 23294 Answers: 1 Comments: 2

Question Number 23290 Answers: 0 Comments: 1

Question Number 23288 Answers: 1 Comments: 3

A uniform chain of mass M and length L is hanging from the table. The chain is in limiting equilibrium when l length of chain over hangs. It is slightly disturbed from this position. Find the speed of the chain just after it completely comes off the table.

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{chain}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{and}\:\mathrm{length} \\ $$$${L}\:\mathrm{is}\:\mathrm{hanging}\:\mathrm{from}\:\mathrm{the}\:\mathrm{table}.\:\mathrm{The}\:\mathrm{chain} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{limiting}\:\mathrm{equilibrium}\:\mathrm{when}\:{l}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{chain}\:\mathrm{over}\:\mathrm{hangs}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{slightly} \\ $$$$\mathrm{disturbed}\:\mathrm{from}\:\mathrm{this}\:\mathrm{position}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chain}\:\mathrm{just}\:\mathrm{after}\:\mathrm{it}\:\mathrm{completely} \\ $$$$\mathrm{comes}\:\mathrm{off}\:\mathrm{the}\:\mathrm{table}. \\ $$

Question Number 23285 Answers: 0 Comments: 0

Mode=17 Mean=10.22 Median=10 What is the shape of the distribution?

$${Mode}=\mathrm{17} \\ $$$${Mean}=\mathrm{10}.\mathrm{22} \\ $$$${Median}=\mathrm{10} \\ $$$$ \\ $$$${What}\:{is}\:{the}\:{shape}\:{of}\:{the}\:{distribution}? \\ $$

Pg 1805 Pg 1806 Pg 1807 Pg 1808 Pg 1809 Pg 1810 Pg 1811 Pg 1812 Pg 1813 Pg 1814

Contact: info@tinkutara.com