aqueous neutral
aqueous acidic
both aqueous acidic and neutral
both aqueous acidic and faintly alkaline
The reaction of permanganate \( (\text{MnO}_4^-) \) with thiosulphate \( (\text{S}_2\text{O}_3^{2-}) \) depends on the pH of the medium. Permanganate can be reduced to different manganese species depending on the conditions.
In a neutral or weakly alkaline medium, permanganate is reduced to manganese(IV) oxide \( (\text{MnO}_2) \):
\( \text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3e^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)} \)
In this reaction, the oxidation state of manganese changes from +7 in \( \text{MnO}_4^- \) to +4 in \( \text{MnO}_2 \), a change of 3 units.
In an acidic medium, permanganate is reduced to manganese(II) ions \( (\text{Mn}^{2+}) \):
\( \text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5e^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)} \)
Here, the oxidation state of manganese changes from +7 to +2, a change of 5 units.
Since the question specifies a change in oxidation state of 3, the reaction must be occurring in a neutral or weakly alkaline medium.
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.

Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: