Question and Answers Forum

OthersQuestion and Answers: Page 131

Question Number 23492 Answers: 0 Comments: 5

Question Number 23489 Answers: 1 Comments: 2

A rectangular wire frame ABCD is in vertical plane is moving with a constant acceleration a into the plane. Direction of gravity is shown in figure. A collar can move on wire AC of length l. Coefficient of friction between wire and collar is μ. Find (i) The minimum acceleration a so that collar does not slip on wire. (ii) The time taken by collar to reach C if acceleration is half the value calculated in part (i)

$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{wire}\:\mathrm{frame}\:{ABCD}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{vertical}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{acceleration}\:{a}\:\mathrm{into}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{Direction} \\ $$$$\mathrm{of}\:\mathrm{gravity}\:\mathrm{is}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}.\:\mathrm{A}\:\mathrm{collar} \\ $$$$\mathrm{can}\:\mathrm{move}\:\mathrm{on}\:\mathrm{wire}\:{AC}\:\mathrm{of}\:\mathrm{length}\:{l}. \\ $$$$\mathrm{Coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{wire} \\ $$$$\mathrm{and}\:\mathrm{collar}\:\mathrm{is}\:\mu.\:\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{acceleration}\:{a}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{collar}\:\mathrm{does}\:\mathrm{not}\:\mathrm{slip}\:\mathrm{on}\:\mathrm{wire}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{collar}\:\mathrm{to}\:\mathrm{reach}\:{C} \\ $$$$\mathrm{if}\:\mathrm{acceleration}\:\mathrm{is}\:\mathrm{half}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{calculated}\:\mathrm{in}\:\mathrm{part}\:\left(\mathrm{i}\right) \\ $$

Question Number 23488 Answers: 0 Comments: 0

area of a(1−cos θ)

$${area}\:{of}\:{a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$

Question Number 23481 Answers: 0 Comments: 5

Which of the diagrams represents variation of total mechanical energy of a pendulum oscillating in air as function of time?

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagrams}\:\mathrm{represents} \\ $$$$\mathrm{variation}\:\mathrm{of}\:\mathrm{total}\:\mathrm{mechanical}\:\mathrm{energy}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{pendulum}\:\mathrm{oscillating}\:\mathrm{in}\:\mathrm{air}\:\mathrm{as}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{time}? \\ $$

Question Number 23476 Answers: 0 Comments: 0

Question Number 23472 Answers: 1 Comments: 1

Question Number 23458 Answers: 0 Comments: 1

Question Number 23444 Answers: 0 Comments: 0

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 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 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 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 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 23294 Answers: 1 Comments: 2

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 23274 Answers: 0 Comments: 2

Assertion: Both N_2 and NO^+ are diamagnetic substances. Reason: NO^+ is isoelectronic with N_2 .

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{Both}\:\mathrm{N}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{NO}^{+} \:\mathrm{are} \\ $$$$\mathrm{diamagnetic}\:\mathrm{substances}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{NO}^{+} \:\mathrm{is}\:\mathrm{isoelectronic}\:\mathrm{with}\:\mathrm{N}_{\mathrm{2}} . \\ $$

Question Number 23272 Answers: 1 Comments: 1

Question Number 23270 Answers: 1 Comments: 0

The reaction CH_4 (g) + Cl_2 (g) → CH_3 Cl(g) + HCl(g) has ΔH = −25 kcal and bond dissociation energy of C − Cl, H − Cl, C − H and Cl − Cl is given as 84 kcal/mol, 103 kcal/mol, x kcal/mol and y kcal/mol respectively. Given x : y = 9 : 5, then bond energy of Cl − Cl bond in kcal/mol is

$$\mathrm{The}\:\mathrm{reaction}\:\mathrm{CH}_{\mathrm{4}} \left(\mathrm{g}\right)\:+\:\mathrm{Cl}_{\mathrm{2}} \left(\mathrm{g}\right)\:\rightarrow\:\mathrm{CH}_{\mathrm{3}} \mathrm{Cl}\left(\mathrm{g}\right) \\ $$$$+\:\mathrm{HCl}\left(\mathrm{g}\right)\:\mathrm{has}\:\Delta\mathrm{H}\:=\:−\mathrm{25}\:\mathrm{kcal}\:\mathrm{and}\:\mathrm{bond} \\ $$$$\mathrm{dissociation}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{C}\:−\:\mathrm{Cl},\:\mathrm{H}\:−\:\mathrm{Cl}, \\ $$$$\mathrm{C}\:−\:\mathrm{H}\:\mathrm{and}\:\mathrm{Cl}\:−\:\mathrm{Cl}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as}\:\mathrm{84}\:\mathrm{kcal}/\mathrm{mol}, \\ $$$$\mathrm{103}\:\mathrm{kcal}/\mathrm{mol},\:\mathrm{x}\:\mathrm{kcal}/\mathrm{mol}\:\mathrm{and}\:\mathrm{y}\:\mathrm{kcal}/\mathrm{mol} \\ $$$$\mathrm{respectively}.\:\mathrm{Given}\:\mathrm{x}\::\:\mathrm{y}\:=\:\mathrm{9}\::\:\mathrm{5},\:\mathrm{then} \\ $$$$\mathrm{bond}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{Cl}\:−\:\mathrm{Cl}\:\mathrm{bond}\:\mathrm{in}\:\mathrm{kcal}/\mathrm{mol} \\ $$$$\mathrm{is} \\ $$

Question Number 23237 Answers: 0 Comments: 6

Question Number 23224 Answers: 0 Comments: 4

Consider the system shown in the figure. Initially the system was in rest. (i) Find the acceleration of block if man climbs the rod with acceleration a (w.r.t. rod) (ii) If the man climb to the top of the rod then find the distance moved by the block.

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{system}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{figure}.\:\mathrm{Initially}\:\mathrm{the}\:\mathrm{system}\:\mathrm{was}\:\mathrm{in}\:\mathrm{rest}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{block}\:\mathrm{if}\:\mathrm{man} \\ $$$$\mathrm{climbs}\:\mathrm{the}\:\mathrm{rod}\:\mathrm{with}\:\mathrm{acceleration}\:{a}\:\left(\mathrm{w}.\mathrm{r}.\mathrm{t}.\right. \\ $$$$\left.\mathrm{rod}\right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{man}\:\mathrm{climb}\:\mathrm{to}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{rod}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{moved}\:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{block}. \\ $$

Question Number 23205 Answers: 0 Comments: 0

thank u.....

$${thank}\:{u}..... \\ $$

Question Number 23187 Answers: 0 Comments: 0

plz anyone answer the question 23181...plz plz plz...

$${plz}\:\:\:{anyone}\:{answer}\:{the}\:{question}\: \\ $$$$\mathrm{23181}...{plz}\:{plz}\:{plz}... \\ $$

Question Number 23181 Answers: 0 Comments: 2

show that the curve with parametric equcations x=t^2 −3t+5, y=t^3 +t^2 −10t+9 intersect at the point (3,1).

$${show}\:{that}\:{the}\:{curve}\:{with}\:{parametric}\: \\ $$$${equcations}\:\:{x}={t}^{\mathrm{2}} \:−\mathrm{3}{t}+\mathrm{5}, \\ $$$${y}={t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:−\mathrm{10}{t}+\mathrm{9}\:{intersect}\:{at}\:{the}\: \\ $$$${point}\:\left(\mathrm{3},\mathrm{1}\right). \\ $$

Pg 126 Pg 127 Pg 128 Pg 129 Pg 130 Pg 131 Pg 132 Pg 133 Pg 134 Pg 135

Contact: info@tinkutara.com