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IIT-JEE Physics previous questions based on Bernoulli's... Continuing with our discussion on Bernoulli's theorem, we take a look at some of the previous questions asked in the Physics paper of IIT-JEE dealing with the same.We would also be trying to find a pattern in the theoretical aspects of the said questions. [ Answers in the next post dealing with Bernoulli's theorem ] *1]...

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IIT-JEE PHYSICS 2009 FULLY SOLVED PAPERS / HOW TO SOLVE... Dear Readers, We are excited to announce two new posts to be published shortly...........Get to see the fully solved papers of the Physics section of IIT-JEE 2009.In addition,get to know the techniques needed to solve almost any physics problem..irrespective of whether it's a school level question or that belonging to...

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Question of the Week : IIT-JEE Physics 2009 /Mechanics   If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame,one can surely say that A] linear momentum of the system does not change in time B] kinetic energy of the system does not change in time C] angular momentum of the system does not change in time D] potential...

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Question of the Week : Answer to IIT-JEE Physics 2009/Optics Given below is the answer to Optics question from IIT-JEE Physics paper 2009 V is taken by calculating the root of  2 x 10 x [ 20 - 12.8 ] which gives  12 m/s.... now, Considering  the diffraction at the water surface, v/u2  = u/u1  or  dv/dt  = u2/u1 x dv/dt     i.e,  dv/dt  = 4/3 x 12 = 16 m/s. Hence,option...

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Question of the Week : from IIT-JEE 2009/Physics /... A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant when the ball is 12.8 m above the water surface, the fish sees the speed of the ball as (Take g = 10 ms–2) (a)...

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Notes / QB

Calculus Based Physics Formulas: Mechanics

 

One dimensional Equations of motion (along a single vector direction)
Velocity as a function of time : v_{xf} = v_{xi} + a_x t
Position as a function of time: x_f = x_i + v_{xi}t + \frac{1}{2} a_x t^2
Velocity as a function of position: v^2_{xf} = v^2_{xi} + 2a_x ( x_f - x_i)<br />

Projectile Motion
Horizontal motion
Velocity along x: v_{xi} = v_i cos(\theta)
Position from position as a function of time: x_f = v_i cos(\theta)t
Max Horizontal dist: R = v^2_i \frac{sin( 2 \theta_i)}{g}

Vertical Motion
Velocity along y:v_{yi} = v_i sin(\theta)
Position: from position as a function of timey_f = y_i + v_{yi}t - \frac{1}{2} g*t^2
Maximum Height: h_{max} = v^2_{i} \frac{sin(\theta_i)}{2g}

Circular Motion
Radial Acc: a_r = v^2_r = a cos( \theta)
Tan.

Acc:a_t = \frac{d \mid \vec {v}\mid}{dt} = a sin(\theta)= r \alpha
Total Acc (magnitude) from Pythagoras: a = \sqrt{a_r^2 + a_t^2}

The Laws of Motion
Newtons Second Law: \sum{F_{x,y, or z}} = ma_{x,y, or z}
Equilibrium Conditions: \sum {F_{x,y, or z}} = 0
Force of Static Friction F_{s max} = \mu_s*n
Force from Kinetic Friction F_{k max} = \mu_k*n

Force/Work
Constant Force: w_{net} = \vec{f_{net}}*\delta r = F * r cos(\theta) = \delta K
Variable Force: w_{net} = \int f_{net} d \vec{r}
Hooke’s Law: f_s = -k x
Spring Work: w = \frac{1}{2} k x_i^2 - \frac{1}{2} k x^2_f
Kinetic energy: k = \frac{1}{2} m v^2
Work – kinetic energy theorem: w_{net} = \delta k = k_f - k_i, k_f = k_i
Power: P = \frac {\Delta w}{\Delta t} ,p = \frac {de}{dt} , P = \vec{f} \vec{v}
gravitational potential energy: U = mgh
conservation of mechanical energy: E = K_f + U_f = K_i + U_i = const. + \mid f_k \delta x \mid
elastic collision conserved moment and KE: v_{1f} = (\frac{m_1 - m_2}{m_1 + m_2}) v_{1i} + (\frac {2 m_2}{m_1 + m_2}) v_{2i}
2d elastic (comp): m_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx},m_1 v_{1ix} = m_1 v_{1f} cos(\theta) + m_2 v_{2fx}cos(\phi)
KE conservation for elastic: \frac{1}{2} m_i v_i^2 +\frac{1}{2} m_{2i} v_{2i}^2 = \frac{1}{2} m_1 i v_{1f} i^2 + \frac{1}{2} m_2 i v_{2f} i^2
Momentum: \vec{P} = m \vec{{v}

Mass
Center of mass (comp): x_{cm} = \frac{\sum_{i=1}^{n} m_i x_i }{m}
Position vector for CM: \vec{r_{cm}} = x_{cm} \vec {i} + y_{cm} \vec {j} + z_{cm} \vec {k}
Continuous mas dist: x_{cm} =\frac {1}{m} \int{\lambda dx}
Mass of Uniform: m= \int{\lambda dx}
Linear Mass Dist: \lambda = \frac{m}{l} = \frac{dm}{dl}
Area Mass Dist: \omega = \frac{m}{a} = \frac{dm}{da}

Rotational Motion
angular speed: \omega = \frac{d\theta}{dt}
angular acceleration: \alpha = \frac{d\omega}{dt} ,\frac{a_t}{r},\frac{\tau}{I}
Moment of Inertia: I = m_i r_i^2,I = \int r^2 dm,I = \int (density)r^2 dv,I = I_cm + mD^2
Rotational KE: K_R = \frac{1}{2} I \omega^2
Work: \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2
Net torque : \sum\tau = I \alpha, \sum \tau =\frac {dL}{dt}
Work : W = \int_{\theta_f}^{\theta_i} t d\theta
Power : P = \tau \omega
Angular Momentum : L = I \omega
Torque: : \tau = rF sin(\theta)

Moments of Inertia
Hoop : I_{cm} =mr^2
Cylinder (hollow) : I_{cm} = \frac{1} {2} m(r_1^2 + r_2^2)
Cylinder : I_{cm} = \frac{1} {2} mr^2
Rectangular Plate : I_{cm} = \frac{1} {12} m(a^2 + b^2)
Rod (center rotate): I_{cm} =\frac{1} {12} mL^2
Rod (end Rotate): I_{cm} =\frac{1} {3} mL^2
Solid Sphere : I_{cm} = \frac{2} {5} mr^2
Spherical Shell : I_{cm} = \frac{2} {3} mr^2

 

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