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Question Id: MA 0000002 Question :  The perimeter of a triangle is 36 cm if its sides are in the ratio of 1:3:2 find the all sides of the triangle ? Answer :  \(6cm\), \(18cm\) and \(12cm\) Explanation : Ratio of sides of triangle is given 1:3:2 Let, sides are \(x\), \(3x\) and \(2x\) Then perimeter will be, \(x+2x+3x=6x\) According to question perimeter is given 36cm. So  \(6x=36\) \[\Rightarrow x=\frac{36}{6}\] \[\Rightarrow x=6\] So,  now \(x=6\), \(3x=18\) and \(2x=12\) Hence, sides of the triangle are 6cm, 18cm and 12cm. (Answer)                          

Question Id: MA 0000001 Question :  Find the roots of the equation \(4-11x=3^{2}\) To find : roots of the equation \(4-11x=3x^{2}\) Solution :   Given a quadratic equation , so it must have two roots \[4-11x=3x^{2}\] \[\Rightarrow 3x^{2}+11x-4=0\] \[\Rightarrow 3x^{2}+12x-x-4=0\] \[\Rightarrow 3x(x+4)-1(x+4)=0\] \[\Rightarrow (3x-1)(x+4)=0\] So, either \(\Rightarrow 3x-1=0\) or \(x+4=0\) If \(\Rightarrow 3x-1=0\) then \(x=\frac{1}{3}\) If \(\Rightarrow x+4=0\) then \(x=-4\) Hence, roots of the equation are \(-4\) and \(-\frac{1}{3}\) . (Answer)   If you have any problem regarding this please leave a comment 💬  below

Question Id:  PH0000004 Question :  What is the Newton's formula for spherical mirror? Solution :   A useful formula for spherical mirror is Newton's formula Here "O" is the object and "I" is the image of the object. Let the distance of the object from the pole is \(u\) and distance of the object from the pole is \(v\) .  Let the distance of the object and the image from the principal focus be \(x\) and \(y\) respectively. So, \(x=u-f\Rightarrow u=f+x\) And \(y=v-f\Rightarrow v=f+y\) Using the value of \(u\) and \(v\) in the formula of mirror gives \[\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\] \[\Rightarrow \frac{1}{f+x}+\frac{1}{f+y}=\frac{1}{f}\] \[\Rightarrow \frac{f+y+f+x}{(f+x)(f+y)}=\frac{1}{f}\] \[\Rightarrow \frac{2f+y+x}{(f+x)(f+y)}=\frac{1}{f}\] \[\Rightarrow 2f^{2}+fy+fx=f^{2}+fx+fy+xy\] \[\Rightarrow 2f^{2}=f^{2}+xy\] \[\Rightarrow f^{2}=xy\] The above equation is the Newton's formula for Spherical mirror. As \(f\) is constant [ i.e \(f=\frac{r}{2}...

Question Id:  PH0000003 Question :  Which of the following equation correctly represents the momentum of a photon of energy E? (a) \(\frac{E}{c}\) (b)  \(E^{2}c\)  (c)  \(Ec\)  (d)  \(Ec^{2}\) Solution : Here, option (a) \(\frac{E}{c}\) is the correct answer. Explanation : Energy carried by a photon having frequency \(\nu\) is  \(E=h\nu\) ; Where \(h\) is Plank's constant. \(\Rightarrow E=\frac{hc}{\lambda}\) \(\Rightarrow \lambda=\frac{hc}{E}\) ; Where \(\lambda\) is wavelength and \(c\) is velocity of light. We know the de Broglie wavelength \(\lambda=\frac{h}{p}\)   \[\Rightarrow p=\frac{h}{\lambda}\] \(\Rightarrow p=\frac{h}{\frac{hc}{E}}\) ; putting the value of \(\lambda\) \[\Rightarrow p=\frac{hE}{hc}\] \[\Rightarrow p=\frac{E}{c}\] So, this is the equation of momentum of a photon.   If you have any problem regarding this please leave a comment 💬  below

Question Id: PH0000002 Question :   Find radius of the curvature of the concave mirror having focal length 7.5 cm. To find : radius (r) of the curvature Solution :   we know the relation between focal length (f) and radius (r)  of curvature of mirror is \[\Rightarrow f=\frac{r}{2}\] \[\Rightarrow r=2 \times f\] \[\Rightarrow r=2 \times 7.5\] \[\Rightarrow r=15\] ∴ Hence radius of the curvature of the concave mirror is \(15cm\) . (Answer)   If you have any problem regarding this please leave a comment 💬 below

Question Id: PH0000001 Question :  Define and give mathematical expression for absolute error, relative error and percentage error. Solution :   Absolute Error: The difference between the measured or inferred value of a quantity and its actual value is called the absolute error ( \(AE\)  ).  So, \(AE=V_{a}-V_{e}\)   Where \(V_{a}\)  is approximate value and \(V_{e}\) is the exact value. Relative Error: The ratio of the absolute error of the measurement to the actual measurement is called relative error ( \(RE\)   ). So, \(RE=\frac{AE}{V_{e}}=\left | \frac{V_{a}-V_{e}}{V_{e}} \right |\) Percent error( \(PE\) ) is the difference between estimated value and the actual value in comparison to the actual value and is expressed as a percentage. In other words, the percent error is the relative error multiplied by 100. So, \(PE=R_{e}\times 100%=^{}\frac{AE}{V_{e}}=\left | \frac{V_{a}-V_{e}}{V_{e}} \right |\times 100%\)   ∴ Above are the definiti...