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Question Number 16504 Answers: 2 Comments: 0

An object moves in a circular path with a constant speed in the xy plane with the centre at the origin. When the object is at x = −2 m, its velocity is −(4 m/s)j^∧ . Then objects velocity at y = 2 m is (1) 4 m/s i^∧ (2) (−4 m/s) i^∧ Using this data, find objects acceleration when it is at y = 2 m (1) 8 m/s^2 i^∧ (2) −8 m/s^2 j^∧

$$\mathrm{An}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{is}\:\mathrm{at}\:{x}\:=\:−\mathrm{2}\:\mathrm{m},\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$$−\left(\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\overset{\wedge} {{j}}.\:\mathrm{Then}\:\mathrm{objects}\:\mathrm{velocity}\:\mathrm{at} \\ $$$${y}\:=\:\mathrm{2}\:\mathrm{m}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}\:\mathrm{m}/\mathrm{s}\:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{i}} \\ $$$$\mathrm{Using}\:\mathrm{this}\:\mathrm{data},\:\mathrm{find}\:\mathrm{objects}\:\mathrm{acceleration} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:{y}\:=\:\mathrm{2}\:\mathrm{m} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{j}} \\ $$

Question Number 16499 Answers: 1 Comments: 0

A particle is moving in parabolic path x^2 = y, with constant speed u. Find the acceleration of the particle when it crossess origin. Also find the radius of curvature at origin.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{parabolic}\:\mathrm{path} \\ $$$${x}^{\mathrm{2}} \:=\:{y},\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{u}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it} \\ $$$$\mathrm{crossess}\:\mathrm{origin}.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of} \\ $$$$\mathrm{curvature}\:\mathrm{at}\:\mathrm{origin}. \\ $$

Question Number 16491 Answers: 0 Comments: 0

Men are running in a line along a road with velocity 9 km/hr behind one another at equal distances of 20 m. Cyclists are also riding along the same line in the same direction at 18 km/hr at equal intervals of 30 m. The speed with which an observer must travel along the road in opposite direction of so that whenever he meets a runner he also meets a cyclist is (1) 9 km/h (2) 12 km/h (3) 18 km/h (4) 6 km/h

$$\mathrm{Men}\:\mathrm{are}\:\mathrm{running}\:\mathrm{in}\:\mathrm{a}\:\mathrm{line}\:\mathrm{along}\:\mathrm{a}\:\mathrm{road} \\ $$$$\mathrm{with}\:\mathrm{velocity}\:\mathrm{9}\:\mathrm{km}/\mathrm{hr}\:\mathrm{behind}\:\mathrm{one} \\ $$$$\mathrm{another}\:\mathrm{at}\:\mathrm{equal}\:\mathrm{distances}\:\mathrm{of}\:\mathrm{20}\:\mathrm{m}. \\ $$$$\mathrm{Cyclists}\:\mathrm{are}\:\mathrm{also}\:\mathrm{riding}\:\mathrm{along}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{line}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{at}\:\mathrm{18}\:\mathrm{km}/\mathrm{hr} \\ $$$$\mathrm{at}\:\mathrm{equal}\:\mathrm{intervals}\:\mathrm{of}\:\mathrm{30}\:\mathrm{m}.\:\mathrm{The}\:\mathrm{speed} \\ $$$$\mathrm{with}\:\mathrm{which}\:\mathrm{an}\:\mathrm{observer}\:\mathrm{must}\:\mathrm{travel} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{road}\:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}\:\mathrm{of} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{whenever}\:\mathrm{he}\:\mathrm{meets}\:\mathrm{a}\:\mathrm{runner}\:\mathrm{he} \\ $$$$\mathrm{also}\:\mathrm{meets}\:\mathrm{a}\:\mathrm{cyclist}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{9}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{12}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{18}\:\mathrm{km}/\mathrm{h} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{6}\:\mathrm{km}/\mathrm{h} \\ $$

Question Number 16497 Answers: 1 Comments: 0

Two particles are revolving on two coplanar circles with constant angular velocities ω_1 and ω_2 respectively. Their time periods are T_1 and T_2 then prove that the time taken by second particle to complete one revolution more than the first particle, T, is given by T = ((T_1 T_2 )/(T_1 − T_2 ))

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{are}\:\mathrm{revolving}\:\mathrm{on}\:\mathrm{two} \\ $$$$\mathrm{coplanar}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular} \\ $$$$\mathrm{velocities}\:\omega_{\mathrm{1}} \:\mathrm{and}\:\omega_{\mathrm{2}} \:\mathrm{respectively}.\:\mathrm{Their} \\ $$$$\mathrm{time}\:\mathrm{periods}\:\mathrm{are}\:{T}_{\mathrm{1}} \:\mathrm{and}\:{T}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{second}\:\mathrm{particle} \\ $$$$\mathrm{to}\:\mathrm{complete}\:\mathrm{one}\:\mathrm{revolution}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{particle},\:{T},\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${T}\:=\:\frac{{T}_{\mathrm{1}} {T}_{\mathrm{2}} }{{T}_{\mathrm{1}} \:−\:{T}_{\mathrm{2}} } \\ $$

Question Number 16466 Answers: 1 Comments: 0

Question Number 16455 Answers: 2 Comments: 0

The speed of a projectile when it is at its greatest height is (√(2/5)) times its speed at half the maximum height. What is its angle of projection?

$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{its}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{is}\:\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\:\mathrm{times}\:\mathrm{its} \\ $$$$\mathrm{speed}\:\mathrm{at}\:\mathrm{half}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{its}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}? \\ $$

Question Number 16310 Answers: 1 Comments: 3

Question Number 16281 Answers: 0 Comments: 0

A plane moves in windy weather due east while the pilot points the plane somewhat south of east. The wind is blowing at 50 km/hr directed 30° east of north, while the plane moves at 200 km/hr relative to the wind. What is the velocity of the plane relative to the ground and what is the direction in which the pilot points the plane?

Question Number 16271 Answers: 1 Comments: 0

The unemployment rate among workers under 25 in a state went from 8.2% to 7.5% in one year. Assume an average of 1340200 workers and estimate the decrease in the number unemployed.

$${The}\:{unemployment}\:{rate}\:{among}\:{workers} \\ $$$${under}\:\mathrm{25}\:{in}\:{a}\:{state}\:{went}\:{from}\:\mathrm{8}.\mathrm{2\%} \\ $$$${to}\:\mathrm{7}.\mathrm{5\%}\:{in}\:{one}\:{year}.\:{Assume}\:{an}\:{average} \\ $$$${of}\:\mathrm{1340200}\:{workers}\:{and}\:{estimate} \\ $$$${the}\:{decrease}\:{in}\:{the}\:{number}\:{unemployed}. \\ $$

Question Number 16616 Answers: 1 Comments: 1

A particle starts from the origin with velocity (√(44)) ms^(−1) on a straight horizontal road. Its acceleration varies with displacement as shown. The velocity of the particle as it passes through the position x = 0.2 km is [Answer: 18 ms^(−1) ]

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{with} \\ $$$$\mathrm{velocity}\:\sqrt{\mathrm{44}}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{on}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{horizontal}\:\mathrm{road}.\:\mathrm{Its}\:\mathrm{acceleration}\:\mathrm{varies} \\ $$$$\mathrm{with}\:\mathrm{displacement}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{The} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{as}\:\mathrm{it}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{position}\:{x}\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{km}\:\mathrm{is} \\ $$$$\left[\mathrm{Answer}:\:\mathrm{18}\:\mathrm{ms}^{−\mathrm{1}} \right] \\ $$

Question Number 16156 Answers: 0 Comments: 2

A body in a uniform horizontal circular motion possesses a variable velocity. Does it mean that the K.E. of the body is also variable?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{in}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{horizontal} \\ $$$$\mathrm{circular}\:\mathrm{motion}\:\mathrm{possesses}\:\mathrm{a}\:\mathrm{variable} \\ $$$$\mathrm{velocity}.\:\mathrm{Does}\:\mathrm{it}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{the}\:\mathrm{K}.\mathrm{E}. \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{is}\:\mathrm{also}\:\mathrm{variable}? \\ $$

Question Number 16155 Answers: 1 Comments: 0

A body of mass m is projected with a speed v making an angle θ with the vertical. What is the change in momentum of the body along the Y- axis; between the starting point and the highest point of its path?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{speed}\:{v}\:\mathrm{making}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{vertical}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{change}\:\mathrm{in} \\ $$$$\mathrm{momentum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{along}\:\mathrm{the}\:\mathrm{Y}- \\ $$$$\mathrm{axis};\:\mathrm{between}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{highest}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}? \\ $$

Question Number 16152 Answers: 1 Comments: 0

In long jump, does it matter how high you jump? What factors determine the span of the jump?

$$\mathrm{In}\:\mathrm{long}\:\mathrm{jump},\:\mathrm{does}\:\mathrm{it}\:\mathrm{matter}\:\mathrm{how}\:\mathrm{high} \\ $$$$\mathrm{you}\:\mathrm{jump}?\:\mathrm{What}\:\mathrm{factors}\:\mathrm{determine}\:\mathrm{the} \\ $$$$\mathrm{span}\:\mathrm{of}\:\mathrm{the}\:\mathrm{jump}? \\ $$

Question Number 16150 Answers: 1 Comments: 1

A projectile is fired at an angle θ with the horizontal direction from O. Neglecting the air friction, it hits the ground at B after 3 seconds. What is the height of point A from ground? [Use g = 10 m/s^2 ]

$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{horizontal}\:\mathrm{direction}\:\mathrm{from}\:{O}. \\ $$$$\mathrm{Neglecting}\:\mathrm{the}\:\mathrm{air}\:\mathrm{friction},\:\mathrm{it}\:\mathrm{hits}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{at}\:{B}\:\mathrm{after}\:\mathrm{3}\:\mathrm{seconds}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{from}\:\mathrm{ground}? \\ $$$$\left[\mathrm{Use}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$

Question Number 16139 Answers: 2 Comments: 0

Path of the bomb released from an aeroplane moving with uniform velocity at certain height as observed by the pilot is (a) a straight line (b) a parabola (c) a circle (d) none of the above

$$\mathrm{Path}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bomb}\:\mathrm{released}\:\mathrm{from}\:\mathrm{an} \\ $$$$\mathrm{aeroplane}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{uniform} \\ $$$$\mathrm{velocity}\:\mathrm{at}\:\mathrm{certain}\:\mathrm{height}\:\mathrm{as}\:\mathrm{observed} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{pilot}\:\mathrm{is} \\ $$$$\left({a}\right)\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\left({b}\right)\:\mathrm{a}\:\mathrm{parabola} \\ $$$$\left({c}\right)\:\mathrm{a}\:\mathrm{circle} \\ $$$$\left({d}\right)\:\mathrm{none}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above} \\ $$

Question Number 16138 Answers: 0 Comments: 0

How many nodal planes are present in 4d_z^2 ?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{nodal}\:\mathrm{planes}\:\mathrm{are}\:\mathrm{present}\:\mathrm{in} \\ $$$$\mathrm{4d}_{\mathrm{z}^{\mathrm{2}} } \:? \\ $$

Question Number 16137 Answers: 0 Comments: 0

For 2s orbital Ψ_r = (1/(√8))((z/a_0 ))^(3/2) (2 − ((zr)/a_0 ))e^(−((zr)/(2a_0 ))) then, hydrogen radial node will be at the distance of (1) a_0 (2) 2a_0 (3) (a_0 /2) (4) (a_0 /3)

$$\mathrm{For}\:\mathrm{2}{s}\:\mathrm{orbital}\:\Psi_{\mathrm{r}} \:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{8}}}\left(\frac{\mathrm{z}}{\mathrm{a}_{\mathrm{0}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{2}\:−\:\frac{\mathrm{zr}}{\mathrm{a}_{\mathrm{0}} }\right)\mathrm{e}^{−\frac{\mathrm{zr}}{\mathrm{2a}_{\mathrm{0}} }} \\ $$$$\mathrm{then},\:\mathrm{hydrogen}\:\mathrm{radial}\:\mathrm{node}\:\mathrm{will}\:\mathrm{be}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{a}_{\mathrm{0}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2a}_{\mathrm{0}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{3}} \\ $$

Question Number 16136 Answers: 1 Comments: 0

Photoelectric emission is observed from a surface when lights of frequency n_1 and n_2 incident. If the ratio of maximum kinetic energy in two cases is K : 1 then (Assume n_1 > n_2 ) threshold frequency is (1) (K − 1) × (Kn_2 − n_1 ) (2) ((Kn_1 − n_2 )/(1 − K)) (3) ((K − 1)/(Kn_1 − n_2 )) (4) ((Kn_2 − n_1 )/(K − 1))

$$\mathrm{Photoelectric}\:\mathrm{emission}\:\mathrm{is}\:\mathrm{observed}\:\mathrm{from} \\ $$$$\mathrm{a}\:\mathrm{surface}\:\mathrm{when}\:\mathrm{lights}\:\mathrm{of}\:\mathrm{frequency}\:{n}_{\mathrm{1}} \\ $$$$\mathrm{and}\:{n}_{\mathrm{2}} \:\mathrm{incident}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{maximum} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{two}\:\mathrm{cases}\:\mathrm{is}\:\mathrm{K}\::\:\mathrm{1} \\ $$$$\mathrm{then}\:\left(\mathrm{Assume}\:{n}_{\mathrm{1}} \:>\:{n}_{\mathrm{2}} \right)\:\mathrm{threshold} \\ $$$$\mathrm{frequency}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{K}\:−\:\mathrm{1}\right)\:×\:\left(\mathrm{K}{n}_{\mathrm{2}} \:−\:{n}_{\mathrm{1}} \right) \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{K}{n}_{\mathrm{1}} \:−\:{n}_{\mathrm{2}} }{\mathrm{1}\:−\:\mathrm{K}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{K}\:−\:\mathrm{1}}{\mathrm{K}{n}_{\mathrm{1}} \:−\:{n}_{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{K}{n}_{\mathrm{2}} \:−\:{n}_{\mathrm{1}} }{\mathrm{K}\:−\:\mathrm{1}} \\ $$

Question Number 16135 Answers: 0 Comments: 0

An electron is moving in 3^(rd) orbit of Hydrogen atom. The frequency of moving electron is (1) 2.19 × 10^(14) rps (2) 7.3 × 10^(14) rps (3) 2.44 × 10^(14) rps (4) 7.3 × 10^(10) rps

$$\mathrm{An}\:\mathrm{electron}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{orbit}\:\mathrm{of} \\ $$$$\mathrm{Hydrogen}\:\mathrm{atom}.\:\mathrm{The}\:\mathrm{frequency}\:\mathrm{of} \\ $$$$\mathrm{moving}\:\mathrm{electron}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}.\mathrm{19}\:×\:\mathrm{10}^{\mathrm{14}} \:\mathrm{rps} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{7}.\mathrm{3}\:×\:\mathrm{10}^{\mathrm{14}} \:\mathrm{rps} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}.\mathrm{44}\:×\:\mathrm{10}^{\mathrm{14}} \:\mathrm{rps} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{7}.\mathrm{3}\:×\:\mathrm{10}^{\mathrm{10}} \:\mathrm{rps} \\ $$

Question Number 16134 Answers: 0 Comments: 0

The mathematical expression which is true for the uncertainty principle is (1) (Δx) (Δv) ≥ (h/(4π)) (2) (ΔE) (Δx) ≥ (h/(4π)) (3) (Δθ) (Δφ) ≥ (h/(4π)) (4) (Δx) (Δm) ≥ (h/(4π))

Question Number 16133 Answers: 1 Comments: 0

H_α line of Balmer series is 6500 A^o . The wave length of Hγ is (1) 4815 A^o (2) 4298 A^o (3) 7800 A^o (4) 3800 A^o

$$\mathrm{H}_{\alpha} \:\mathrm{line}\:\mathrm{of}\:\mathrm{Balmer}\:\mathrm{series}\:\mathrm{is}\:\mathrm{6500}\:\overset{\mathrm{o}} {\mathrm{A}}.\:\mathrm{The} \\ $$$$\mathrm{wave}\:\mathrm{length}\:\mathrm{of}\:\mathrm{H}\gamma\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4815}\:\overset{\mathrm{o}} {\mathrm{A}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{4298}\:\overset{\mathrm{o}} {\mathrm{A}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{7800}\:\overset{\mathrm{o}} {\mathrm{A}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{3800}\:\overset{\mathrm{o}} {\mathrm{A}} \\ $$

Question Number 16102 Answers: 0 Comments: 0

Question Number 16081 Answers: 2 Comments: 0

A spherical balloon of 21 cm diameter is to be filled with hydrogen at NTP from a cylinder containing the gas at 20 atmosphere at 27°C. If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up.

$$\mathrm{A}\:\mathrm{spherical}\:\mathrm{balloon}\:\mathrm{of}\:\mathrm{21}\:\mathrm{cm}\:\mathrm{diameter} \\ $$$$\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{with}\:\mathrm{hydrogen}\:\mathrm{at}\:\mathrm{NTP} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{cylinder}\:\mathrm{containing}\:\mathrm{the}\:\mathrm{gas}\:\mathrm{at} \\ $$$$\mathrm{20}\:\mathrm{atmosphere}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{cylinder} \\ $$$$\mathrm{can}\:\mathrm{hold}\:\mathrm{2}.\mathrm{82}\:\mathrm{litres}\:\mathrm{of}\:\mathrm{water},\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{balloons}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{filled}\:\mathrm{up}. \\ $$

Question Number 16044 Answers: 1 Comments: 1

A particle is projected horizontally with speed u from point A, which is 10 m above the ground. If the particle hits the inclined plane perpendicularly at point B. [g = 10 m/s^2 ] Find horizontal speed with which the particle was projected.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{horizontally} \\ $$$$\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{point}\:{A},\:\mathrm{which}\:\mathrm{is}\:\mathrm{10} \\ $$$$\mathrm{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{perpendicularly}\:\mathrm{at} \\ $$$$\mathrm{point}\:{B}.\:\left[{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$$$\mathrm{Find}\:\mathrm{horizontal}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{was}\:\mathrm{projected}. \\ $$

Question Number 16079 Answers: 1 Comments: 1

An open flask contains air at 27°C. To what temperature it must be heated to expel one-fourth of the air?

$$\mathrm{An}\:\mathrm{open}\:\mathrm{flask}\:\mathrm{contains}\:\mathrm{air}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}.\:\mathrm{To} \\ $$$$\mathrm{what}\:\mathrm{temperature}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{heated}\:\mathrm{to} \\ $$$$\mathrm{expel}\:\mathrm{one}-\mathrm{fourth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{air}? \\ $$

Question Number 16032 Answers: 0 Comments: 0

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