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A block of mass m is attached with a massless spring of force constant k. The block is placed over a fixed rough inclined surface for which the coefficient of friction is mu = 3/4. The block of mass
04
Oct
A block of mass m is attached with a massless spring of force constant k. The block is placed over a fixed rough inclined surface for which the coefficient of friction is mu = 3/4. The block of mass A block of mass m is attached with a massless spring of force constant k. The [...]
Find the sum of the following APs. 0.6, 1.7, 2.8 ,…….., to 100 terms
04
Oct
Find the sum of the following APs. 0.6, 1.7, 2.8 ,…….., to 100 terms ........ 1.7 2.8 Find the sum of the following APs. 0.6 to 100 terms October 4, 2020 Category: Chapter 5 - Arithmetic Progressions , Maths , NCERT Class 10 ,
Find the sum of the following APs. − 37, − 33, − 29 ,…, to 12 terms
04
Oct
Find the sum of the following APs. − 37, − 33, − 29 ,…, to 12 terms ... − 29 − 33 Find the sum of the following APs. − 37 to 12 terms October 4, 2020 Category: Chapter 5 - Arithmetic Progressions , Maths , NCERT Class 10 ,
A particle is projected with a velocity u making an angle theta with the horizontal. The instantaneous power of the gravitational force
04
Oct
A particle is projected with a velocity u making an angle theta with the horizontal. The instantaneous power of the gravitational force A particle is projected with a velocity u making an angle theta with the horizontal. The instantaneous power of the gravitational force October 4, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma [...]
Find the sum of the following APs. (i) 2, 7, 12 ,…., to 10 terms.
04
Oct
Find the sum of the following APs. (i) 2, 7, 12 ,…., to 10 terms. -12 .... 7) Find the sum of the following APs. (i) 2 to 10 terms. October 4, 2020 Category: Chapter 5 - Arithmetic Progressions , Maths , NCERT Class 10 ,
The potential energy for a force field F is given by U (x, y) = cos (x + y). The force acting on a particle at position given by coordinates (0, pie/4) is
04
Oct
The potential energy for a force field F is given by U (x, y) = cos (x + y). The force acting on a particle at position given by coordinates (0, pie/4) is pie/4) is The potential energy for a force field F is given by U (x y) = cos (x + y). The [...]
A vessel of volume V = 30 litre is separated into three equal parts by stationary semi permeable membrane . The left , middle and right parts are filled with mH = 30 gm of hydrogen , m O2 = 160 gm of oxygen and mN=70 gm of nitrogen respectively . The left partition lets through only hydrogen while the right partition lets through hydrogen and nitrogen . If the temperature in all is 300 K , the ratio of pressure in the three compartments will be :
04
Oct
A vessel of volume V = 30 litre is separated into three equal parts by stationary semi permeable membrane . The left , middle and right parts are filled with mH = 30 gm of hydrogen , m O2 = 160 gm of oxygen and mN=70 gm of nitrogen respectively . The left partition lets [...]
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A vessel of volume V = 30 litre is separated into three equal parts by stationary semi permeable membrane . The left ,
middle and right parts are filled with mH = 30 gm of hydrogen ,
The rms speed of a molecules of a gas in a vessel is 400 m/s. If half of the gas leaks out at constanmt temperature ,
the rms speed of the remaining molecules will be........... ,
P – T graph for smae number of moles of two ideal gases are shown. Find the path along which volume decreases.
04
Oct
P – T graph for smae number of moles of two ideal gases are shown. Find the path along which volume decreases. P - T graph for smae number of moles of two ideal gases are shown. Find the path along which volume decreases. October 4, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma [...]
The potential energy function associated with the force F = 4 xyi + 2 x^2 j is
04
Oct
The potential energy function associated with the force F = 4 xyi + 2 x^2 j is The potential energy function associated with the force F = 4 xyi + 2 x^2 j is October 4, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lower to its equilibrium position. If S is the gain in spring energy
04
Oct
One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lower to its equilibrium position. If S is the gain in spring energy One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end [...]